aa r X i v : . [ c s . A I] S e p PT 1
Problem Theory
Ram´on Casares orcid:
The Turing machine, as it was presented by Turing himself, modelsthe calculations done by a person. This means that we can com-pute whatever any Turing machine can compute, and therefore weare Turing complete. The question addressed here is why,
Whyare we Turing complete?
Being Turing complete also means thatsomehow our brain implements the function that a universal Turingmachine implements. The point is that evolution achieved Turingcompleteness, and then the explanation should be evolutionary, butour explanation is mathematical. The trick is to introduce a mathe-matical theory of problems, under the basic assumption that solvingmore problems provides more survival opportunities. So we builda problem theory by fusing set and computing theories. Then weconstruct a series of resolvers, where each resolver is defined by itscomputing capacity, that exhibits the following property: all prob-lems solved by a resolver are also solved by the next resolver in theseries if certain condition is satisfied. The last of the conditionsis to be Turing complete. This series defines a resolvers hierarchythat could be seen as a framework for the evolution of cognition.Then the answer to our question would be: to solve most problems .By the way, the problem theory defines adaptation, perception, andlearning, and it shows that there are just three ways to resolve anyproblem: routine, trial, and analogy. And, most importantly, thistheory demonstrates how problems can be used to found mathe-matics and computing on biology.
Keywords: problem solving; adaptation, perception & learning; Turingcompleteness; resolvers hierarchy; evolution of cognition.
This is arXiv:1412.1044 version , and it is licensed as cc-by .Any comments on it to [email protected] are welcome. ww.ramoncasares.com
PT 2
Problem Theory § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § References . . . . . . . . . . . . . . . . . . . . . . . . . 44 ww.ramoncasares.com
PT 3 § Devoid of problems, thinking is useless.
Warning
This paper does not explain how to solve, nor how to resolve, any problem. § ¶ · The object of this paper is to present a mathematical theory of problems. Theresulting problem theory provides meaning to set theory and to computing theory. ¶ · Problems are nearly everywhere. We can say that mathematics is all about mathe-matical problems, but also that physics is all about physical problems, and philosophy isall about philosophical problems. I said nearly because there are not problems in a river;a river just flows. So, where are problems? ¶ · This problem theory gives an answer: There are problems where there is freedom.Determinists will surely object, but they should note that if there were only uncertainty,and not the possibility of doing otherwise, then problem resolving would be purposelessand absurd. Nevertheless, in this theory freedom cannot exist by itself, but freedom isalways limited by a condition and both together, freedom and a condition, are a problem.In fact, the resolution of any problem is the process of spending all of its freedom whilestill satisfying the condition. So resolving is fighting freedom away. And, if people fightfor freedom, it is because we want problems; in fact, not having any problem is boring.But I would say more, we are devices exquisitely selected to resolve problems, becausesurviving is literally the problem of being dead or alive: “To be, or not to be—that is thequestion.” ¶ · I am digressing, sorry! The point is that problems are related to sets at the verybottom: for each problem there is a condition that determines if anything is a solutionto it or not, so for each problem there is a set, the set of its solutions, and the conditionis its characteristic function. This means that problems and sets are just two names forthe same thing. So problem theory, being just a rewording of set theory, would be abetter foundation for mathematics than set theory, because problems are more related tothinking than sets are. ¶ · We have just seen how problems and solutions fit with sets, but we have seen nothingabout resolutions, that is, the ways to go from a problem to its solutions. It is a factthat computing is helping us in resolving many problems. Perhaps too many: How ourmodern society would subsist without computers? I am digressing again, sorry! The rightquestion is: What is the relation between problem resolving and computing? ¶ · Computing is the mechanical manipulation of strings of symbols. Mechanical in thesense that the manipulations do not take into account the meaning of the symbols, butthey just obey blindly a finite set of well-defined rules. Being meaningless, what couldbe the purpose of computing? Historically, computing resulted from two apparentlydifferent pursuits: the foundation of mathematics, and the enhancement of calculatingmachines. The second, the development of mechanical aids to calculation, is easier tounderstand. When we learn the algorithm for division we readily appreciate that thosefixed rules can be better applied by a machine than by a person. This explains why anarithmetic calculator comes handy when resolving a problem that requires performing anumerical division. And it could also help us to understand why computation was seenas the ideal for mathematical rigor, and then how computing relates to the foundationsof mathematics. ww.ramoncasares.com
PT 4 ¶ · But, again, what is the purpose of mathematical formalization? Is it true that acomplete formalization of mathematics would render it meaningless? What would bethe use of something meaningless? And again, the arithmetic calculator, dividing for us,answers the three questions: formalization prevents mistakes and assures that nothinghas been taken for granted, and, while it is literally meaningless, it is not useless if theformalism helps us in resolving problems. Though pending on an if, formalism is not yetlost. This paper cuts that Gordian knot by showing that problem resolving is computing. ¶ · In fact, that ‘resolving is computing’ comes from the founding paper of computing.Turing (1936) proved that the
Entscheidungsproblem , which is the German word for‘decision problem’, is unsolvable, because it cannot be solved by any Turing machine.For this proof to be valid, ‘solved by a Turing machine’ has to be equal to ‘solved’, andtherefore ‘resolved by computing’ has to be redundant. ¶ · Summarizing, a problem is a set, and resolving is computing. This is how this problemtheory relates to set and computing theories at the highest level of abstraction. For amore detailed view you should continue reading this paper. § ¶ · The object of this paper is to introduce a mathematical theory of problems. Becauseour approach is minimalist, aiming to keep only what is essential, we will define a prob-lem theory from first principles. Section § ¶ · Section § § § § § § § ¶ · The next section, Section §
4, is about computing. In Subsection § § § § § ¶ · Section § § ww.ramoncasares.com PT 5 resolver as the set of problems that the resolver resolves. Then we construct a series offive resolvers: ◦ Mechanism, Subsection § ◦ Adapter, in § ◦ Perceiver, in § ◦ Learner, in § ◦ Subject, in § § ¶ · The paper finishes with some conclusions, in Section §
6. In the first subsection, § § § § § ¶ · Every problem is made up of freedom and of a condition. There have to be possibilitiesand freedom to choose among them, because if there is only necessity and fatality, thenthere is neither a problem nor is there a decision to make. The different possible optionscould work, or not, as solutions to the problem, so that in every problem a certain condition that will determine if an option is valid or not as a solution to the problemmust exist. Problem (cid:26)
FreedomCondition ww.ramoncasares.com
PT 6 § ¶ · A fundamental distinction that we must make is between the solution and the res-olution of a problem. Resolving is to searching as solving is to finding, and please notethat one can search for something that does not exist.Resolving · SearchingSolving · FindingThus, resolution is the process that attempts to reach the solutions to the problem, whilea solution of the problem is any use of freedom that satisfies the condition. In the state-transition jargon: a problem is a state of ignorance, a solution is a state of satisfaction,and a resolution is a transition from uncertainty to certainty.Problem Resolution −−−−−−−−−−−−→
Solution ¶ · We can explain this with another analogy. The problem is defined by the tension thatexists between two opposites: freedom, free from any limits, and the condition, which ispure limit. This tension is the cause of the resolution process. But once the condition isfulfilled and freedom is exhausted, the solution annihilates the problem. The resolutionis, then, a process of annihilation that eliminates freedom as well as the condition of theproblem, in order to produce the solution.FreedomCondition | {z }
Problem (cid:27)
Resolution −−−−−−−−−−−−→
Solution ¶ · A mathematical example may also be useful in order to distinguish resolution fromsolution. In a problem of arithmetical calculation, the solution is a number and theresolution is an algorithm such as the algorithm for division, for example. § ¶ · There are three ways to resolve a problem: routine, trial, and analogy.Resolution
RoutineTrialAnalogy ¶ · To resolve a problem by routine , that is, by knowing or without reasoning, it isnecessary to know the solutions, and it is necessary to know that they solve that problem. ¶ · If the solutions to a problem are not known, but it is known a set of possible so-lutions, then we can use a trial and error procedure, that is, we can try the possiblesolutions. To resolve by trial is to test each possible solution until the set of possiblesolutions is exhausted or a halting condition is met. There are two tasks when we try: totest if a particular possibility satisfies the problem condition, and to govern the processdetermining the order of the tests and when to halt. There are several ways to governthe process, that is, there is some freedom in governing the trial, and so, if we also put ww.ramoncasares.com
PT 7 a condition on it, for example a temporal milestone, then governing is a problem. Andthere are three ways to resolve a problem ( da capo ). ¶ · By analogy we mean to transform a problem into a different one, called question,which is usually composed of several subproblems. This works well if the subproblems areeasier to resolve than the original problem. There are usually several ways to transformany problem (there is freedom), but only those transformations that result in questionsthat can be resolved are valid (which is a condition), so applying an analogy to a problemis a problem. There are three ways to resolve the analogy, the question, and each of itssubproblems: routine, trial, and analogy ( da capo ). If we could translate a problem intoan analogue question, and we could find a solution to that question, called answer, andwe could perform the inverse translation on it, then we would have found a solution tothe original problem. Problem Solution ↓ ↑ Question −→ Answer § ¶ · Lastly we are ready to list the eight concepts of the problem theory. They are:problem, with freedom and condition; resolution, with routine, trial, and analogy; andsolution. Problem Theory
Problem (cid:26)
FreedomConditionResolution
RoutineTrialAnalogySolution § § § We will refer to the set of problems as P . We will refer to the set ofresolutions as R . We will refer to the set of solutions as S . Definition
A resolution takes a problem and returns the set of the solutions to theproblem. Then resolutions are R = P → S , where 2 S is the powerset, or the set ofthe subsets, of S . § ⊤ stands for ‘true’, and ⊥ for ‘false’. We will refer to the set of theseBoolean values as B . B = {⊤ , ⊥} . Comment ⊤ = ¬⊥ and ⊥ = ¬⊤ . Also [ P = ⊤ ] = P and [ P = ⊥ ] = ¬ P . § Given s ∈ S ⊆ S and f ∈ F ⊆ ( S → S ), so f : S → S and f ( s ) ∈ S ,we will use the following rewriting rules: f ( S ) = { f ( s ) | s ∈ S } , F ( s ) = { f ( s ) | f ∈ F } , and F ( S ) = { f ( s ) | s ∈ S × f ∈ F } . Comment As f ( s ) ∈ S , then f ( S ) ∈ S , F ( s ) ∈ S , and F ( S ) ∈ S . Proposition If s ∈ S and f ∈ F , then f ( S ) ⊆ F ( S ) and F ( s ) ⊆ F ( S ). ww.ramoncasares.com PT 8 § Problem π is x ? P π ( x ), where P π is any predicate, or Boolean-valuedfunction, on S ; so P π : S → B , where P π ( x ) = ⊤ means that x is a solution of π , and P π ( x ) = ⊥ means that x is not a solution of π . Comment
A problem π = x ? P π ( x ) is made up of freedom and of a condition, as definedin Section §
2. The condition is P π , and freedom is represented by the free variable x ,which is free to take any value in S , x ∈ S . § A function ∗ f is effectively calculable if there is a purely mechanicalprocess to find ∗ f ( s ) for any s . Comment
This definition of effective calculability was stated by Turing (1938), § Comment
If the result of the calculation is finite, then an effective calculation has tocomplete it. If the result of the calculation is infinite, then an effective calculationhas to proceed forever towards the result.
Notation
We will refer to the set of effectively calculable functions as ∗ F . § A problem π is expressible if its condition P π is an effectively cal-culable function. Comment
The result of a condition is in set B = {⊤ , ⊥} , so it is always finite. Thereforea problem is not expressible if for some x we cannot calculate whether x is a solutionor not in a finite time. § The condition isomorphism is the natural isomorphism that relateseach problem π with its condition P π : for each predicate P there is a problem, x ? P ( x ), and for each problem, π = x ? P π ( x ) there is a predicate, P π . That is, P ⇔ ( S → B ) : x ? P π ( x ) ↔ P π . Comment
Using the condition isomorphism, two problems are equal if they have thesame condition, that is, π = ρ ⇔ P π = P ρ . Comment
The condition isomorphism abstracts freedom away. § The set of problems is the set of predicates, that is, P = S → B . Proof P ∼ = S → B , by the condition isomorphism, see § P = S → B . But freedom has to be abstracted away from mathematics becausefreedom is free of form and it cannot be counted nor measured. ⋄ Comment
Although in mathematics we cannot deal with freedom, it is an essentialpart of problems, see § π is its condition P π . § The name of the free variable is not important, it can be replaced: x ? P ( x ) = y ? P ( y ). Proof
By the condition isomorphism, and § x ? P ( x ) and y ? P ( y ),are equal, x ? P ( x ) = y ? P ( y ), because they have the same condition, P . ⋄ Comment
This means that the rule of α -conversion stands for problem expressions.See Curry & Feys (1958), Section 3D. § Let π and ρ be two problems. Then π ∧ ρ = x ? P π ( x ) ∧ P ρ ( x ), and π ∨ ρ = x ? P π ( x ) ∨ P ρ ( x ), and ¯ π = x ? ¬ P π ( x ). Comment
In other words, P π ∧ ρ ( x ) = P π ( x ) ∧ P ρ ( x ), P π ∨ ρ ( x ) = P π ( x ) ∨ P ρ ( x ), and P ¯ π ( x ) = ¬ P π ( x ). Comment
This provides a way to compose, or decompose, problems. ww.ramoncasares.com
PT 9 § A problem τ is tautological if its condition is a tautology; P τ isa tautology, if ∀ x, P τ ( x ) = ⊤ . A problem ¯ τ is contradictory if its condition is acontradiction; P ¯ τ is a contradiction, if ∀ x, P ¯ τ ( x ) = ⊥ . Lemma
Both τ and ¯ τ are expressible. Proof
Because P τ and P ¯ τ are effectively calculable, see § § ⋄§ h P , ∨ , ∧ , ¬ , ¯ τ , τ i is a Boolean algebra, where ¯ τ is the neutral for ∨ ,and τ is the neutral for ∧ . Proof
Because P π ( x ) ∈ B . In detail, ∀ π, ρ, σ ∈ P :1o. ( π ∨ ρ ) ∨ σ = x ? P π ∨ ρ ( x ) ∨ P σ ( x ) = x ? ( P π ( x ) ∨ P ρ ( x )) ∨ P σ ( x ) = x ? P π ( x ) ∨ ( P ρ ( x ) ∨ P σ ( x )) = x ? P π ( x ) ∨ P ρ ∨ σ ( x ) = π ∨ ( ρ ∨ σ ).1a. ( π ∧ ρ ) ∧ σ = x ? P π ∧ ρ ( x ) ∧ P σ ( x ) = x ? ( P π ( x ) ∧ P ρ ( x )) ∧ P σ ( x ) = x ? P π ( x ) ∧ ( P ρ ( x ) ∧ P σ ( x )) = x ? P π ( x ) ∧ P ρ ∧ σ ( x ) = π ∧ ( ρ ∧ σ ).2o. π ∨ ρ = x ? P π ( x ) ∨ P ρ ( x ) = x ? P ρ ( x ) ∨ P π ( x ) = ρ ∨ π .2a. π ∧ ρ = x ? P π ( x ) ∧ P ρ ( x ) = x ? P ρ ( x ) ∧ P π ( x ) = ρ ∧ π .3o. π ∨ ¯ τ = x ? P π ( x ) ∨ P ¯ τ ( x ) = x ? P π ( x ) ∨ ⊥ = x ? P π ( x ) = π .3a. π ∧ τ = x ? P π ( x ) ∧ P τ ( x ) = x ? P π ( x ) ∧ ⊤ = x ? P π ( x ) = π .4o. π ∨ ¯ π = x ? P π ( x ) ∨ P ¯ π ( x ) = x ? P π ( x ) ∨ ¬ P π ( x ) = x ? ⊤ = x ? P τ ( x ) = τ .4a. π ∧ ¯ π = x ? P π ( x ) ∧ P ¯ π ( x ) = x ? P π ( x ) ∧ ¬ P π ( x ) = x ? ⊥ = x ? P ¯ τ ( x ) = ¯ τ .5o. π ∨ ( ρ ∧ σ ) = x ? P π ( x ) ∨ P ρ ∧ σ ( x ) = x ? P π ( x ) ∨ ( P ρ ( x ) ∧ P σ ( x )) = x ? ( P π ( x ) ∨ P ρ ( x )) ∧ ( P π ( x ) ∨ P σ ( x )) = x ? P π ∨ ρ ( x ) ∧ P π ∨ σ ( x ) = ( π ∨ ρ ) ∧ ( π ∨ σ ).5a. π ∧ ( ρ ∨ σ ) = x ? P π ( x ) ∧ P ρ ∨ σ ( x ) = x ? P π ( x ) ∧ ( P ρ ( x ) ∨ P σ ( x )) = x ? ( P π ( x ) ∧ P ρ ( x )) ∨ ( P π ( x ) ∧ P σ ( x )) = x ? P π ∧ ρ ( x ) ∨ P π ∧ σ ( x ) = ( π ∧ ρ ) ∨ ( π ∧ σ ). ⋄ § § Everything is in S . In other words, S is the set of everything. Proof
Anything, let us call it s , is a solution to problem x ? [ x = s ], because equalityis reflexive, and therefore everything satisfies the condition of being equal to itself. ⋄ Comment
Freedom is complete, because x is free to take any value; x ∈ S is not arestriction. And P π : S → B is a predicate on everything. Comment
Some paradoxes derive from this theorem, see § S , see Section §
5. See also Subsection § Corollary P ⊂ S and R ⊂ S . Even B ⊂ S . Comment
If you are a teacher looking for a problem to ask in an exam, then yoursolution is a problem, so P ⊂ S makes sense. And if you are a mathematician lookingfor an algorithm to resolve some kind of problems, then your solution is a resolution,so R ⊂ S makes sense. There are many yes-or-no questions, so B ⊂ S makes sense. § Let Σ π be the (possibly infinite) set of all the solutions to problem π . So Σ π ⊆ S , or Σ π ∈ S , and Σ π = { s | P π ( s ) } . Comment
A solution of the problem is any use of freedom that satisfies the condition,see Section §
2, so s is a solution of problem π , if P π ( s ) stands. Comment
The condition of the problem π is the characteristic function of its set ofsolutions, that is, P π is the characteristic function of Σ π . ww.ramoncasares.com PT 10 § Σ π ∨ ρ = Σ π ∪ Σ ρ , and Σ π ∧ ρ = Σ π ∩ Σ ρ , and Σ ¯ π = Σ π . Proof
Just apply the definitions in § π ∨ ρ = { s | P π ∨ ρ ( s ) } = { s | P π ( s ) ∨ P ρ ( s ) } = { s | s ∈ Σ π ∨ s ∈ Σ ρ } = Σ π ∪ Σ ρ .Σ π ∧ ρ = { s | P π ∧ ρ ( s ) } = { s | P π ( s ) ∧ P ρ ( s ) } = { s | s ∈ Σ π ∧ s ∈ Σ ρ } = Σ π ∩ Σ ρ .Σ ¯ π = { s | P ¯ π ( s ) } = { s | ¬ P π ( s ) } = { s | s / ∈ Σ π } = Σ π . ⋄§ For a tautological problem, x ? P τ ( x ), everything is a solution, Σ τ = S .For a contradictory problem, x ? P ¯ τ ( x ), nothing is a solution, Σ ¯ τ = ∅ . Proof Σ τ = { s | P τ ( s ) } = { s | ⊤ } = S . Σ ¯ τ = { s | P ¯ τ ( s ) } = { s | ⊥ } = {} = ∅ . ⋄§ Σ π ∪ Σ ¯ π = S and Σ π ∩ Σ ¯ π = ∅ . Proof Σ π ∪ Σ ¯ π = { s | P π ( s ) } ∪ { s | ¬ P π ( s ) } = { s | P π ( s ) ∨ ¬ P π ( s ) } = { s | ⊤ } = S .Σ π ∩ Σ ¯ π = { s | P π ( s ) } ∩ { s | ¬ P π ( s ) } = { s | P π ( s ) ∧ ¬ P π ( s ) } = { s | ⊥ } = ∅ . ⋄§ The solutions of π ∧ ρ are solutions of π and of ρ . Proof ∀ s ∈ S ; s ∈ Σ π ∧ ρ ⇔ s ∈ Σ π ∩ Σ ρ ⇔ s ∈ Σ π ∧ s ∈ Σ ρ . ⋄ Comment
The reader is free to explore this Boolean landscape, but here we will closewith the following theorems. § h S , ∪ , ∩ , − , ∅ , S i is a Boolean algebra, where ∅ is the neutral for ∪ ,and S is the neutral for ∩ . Proof
The powerset of a set M , with the operations of union ∪ , intersection ∩ , andcomplement with respect to set M , noted Q , is a typical example of a Boolean algebra.In detail, ∀ Q, R, S ∈ S :1o. ( Q ∪ R ) ∪ S = Q ∪ ( R ∪ S ). 1a. ( Q ∩ R ) ∩ S = Q ∩ ( R ∩ S ).2o. Q ∪ R = R ∪ Q . 2a. Q ∩ R = R ∩ Q .3o. Q ∪ ∅ = Q . 3a. Q ∩ S = Q .4o. Q ∪ Q = S . 4a. Q ∩ Q = ∅ .5o. Q ∪ ( R ∩ S ) = ( Q ∪ R ) ∩ ( Q ∪ S ). 5a. Q ∩ ( R ∪ S ) = ( Q ∩ R ) ∪ ( Q ∩ S ). ⋄§ h P , ∨ , ∧ , ¬ , ¯ τ , τ i is isomorphic to h S , ∪ , ∩ , − , ∅ , S i , that is, P ∼ = 2 S . Proof
We define the bijection Σ that relates each problem π with the set of its solutionsΣ π : for every problem π ∈ P there is a set, the set of its solutions, Σ π ∈ S , andfor every set S ∈ S there is a problem, π S = x ? [ x ∈ S ], where π S ∈ P . Now, byLemma § ∨ ↔ ∪ , ∧ ↔ ∩ , ¬ ↔ − , and, by Lemma § τ ↔ ∅ , τ ↔ S . ⋄ Comment
We will call P ∼ = 2 S the set isomorphism. That is, P ⇔ S : π ↔ Σ π . Comment
Using the set isomorphism, two problems are equal if they have the samesolutions, that is, π = ρ ⇔ Σ π = Σ ρ . § The set of problems is equal to the powerset of the solutions, that is, P = 2 S . Proof
The equality P = 2 S derives directly from the set isomorphism P ∼ = 2 S , see § ⋄§ The set of singletons is: S = { S ∈ S | [ | S | = 1 ] } . Proposition S ⊂ S , because ∀ S ∈ S , S ∈ S , but ∅ ∈ S and ∅ / ∈ S . ww.ramoncasares.com PT 11 § The singleton isomorphism is the isomorphism between S and S that relates each s ∈ S to the set { s } ∈ S , and the converse. That is, S ∼ = S , and S ⇔ S : s ↔ { s } . Comment
We can extend any operation on S to S . For example, for any binaryoperation ∗ on S , we define { a } ∗ { b } = { a ∗ b } . Comment
From the singleton isomorphism: S ∼ = S ⊂ S . § The set of solutions S is a proper subset of the set of problems P , thatis, S ⊂ P . Proof
By the singleton isomorphism, see § S ∼ = S , and, by the set isomorphism,see § S ⊂ P , and then S ∼ = S ⊂ P . ⋄ Paradox
We have both, S ⊂ P and, by § P ⊂ S . Comment
If we only accept computable functions and computable sets, then S ∗ P ∗ ,see Subsection § § A problem π is solved if a solution of π is known. Comment
To solve a problem, given the set of its solutions Σ π , a choice function f c : 2 S \ ∅ → S is needed. § A problem is unsolvable if Σ π = {} = ∅ , that is, if | Σ π | = 0. Aproblem is solvable if | Σ π | > Comment
If a problem has not any solution, then it is unsolvable. If a problem has asolution, then it can be solved. A problem is solvable if it can be solved.
Comment
Solved implies solvable, but not the converse: Solved ⇒ Solvable. § § We will refer to the routine of problem π as R π . The routine is theset of the solutions to the problem, a set that is known, see § R π = Σ π . Comment
The routine of problem π , R π , is then, or an extensive definition of Σ π ,Σ π = { s , . . . , s n } , or a procedure P that generates all problem π solutions and thenhalts. If the number of solutions is infinite, | Σ π | ≥ ℵ , then R π has to be a procedure P that keeps generating solutions forever. § A trial on problem π over the set of possible solutions S , written T π ( S ), returns the set of those elements in S that satisfy the problem condition P π ,see § T π ( S ) = { s ∈ S | P π ( s ) } . Comment
Mathematically we will ignore the practical problem of governing the trial.Practically we will need a halt condition to truncate the calculations that are toolong (or infinite), and some ordering on the tests to fit the execution of the tests tothe available calculating machinery. § To test if a possible solution s ∈ S is a solution to problem π , is toreplace the free variable with s . So, being π = x ? P π ( x ), then to test if s is a solutionis to calculate P π ( s ). Comment
Testing is a calculation S → B . ww.ramoncasares.com PT 12 § Replacing variables in expressions requires not confusing free withbound variables, nor bound with free variables.
Comment
This means that the rule of β -conversion and the rules γ for substitutionstand for testing. See Curry & Feys (1958), Section 3D for β -conversion, and Sec-tion 3E for substitution (the rules γ ). § A trial on problem π over the set S is equal to the intersection of S with the set of the solutions Σ π , that is, T π ( S ) = S ∩ Σ π . Proof T π ( S ) = { s ∈ S | P π ( s ) } = { s | s ∈ S ∧ P π ( s ) } = { s | s ∈ S ∧ s ∈ Σ π } = { s | s ∈ S } ∩ { s | s ∈ Σ π } = S ∩ Σ π . ⋄ Corollary
Any trial is a subset of the set of solutions, T π ( S ) ⊆ Σ π . Proof T π ( S ) = S ∩ Σ π ⊆ Σ π . ⋄ Corollary
Any trial is a subset of the routine, that is, T π ( S ) ⊆ Σ π = R π . § If S is a superset of Σ π , then a trial on problem π over S is equal toΣ π , and the converse, that is, Σ π ⊆ S ⇔ T π ( S ) = Σ π . Proof Σ π ⊆ S ⇔ S ∩ Σ π = Σ π ⇔ T π ( S ) = Σ π , using Theorem § ⋄ Corollary If S is a superset of Σ π , then a trial on problem π over S is equal to theroutine of π , and the converse, that is, Σ π ⊆ S ⇔ T π ( S ) = R π . Proof Σ π ⊆ S ⇔ S ∩ Σ π = Σ π ⇔ T π ( S ) = R π . ⋄ Corollary
A trial on problem π over the whole S is equal to Σ π , that is, T π ( S ) = Σ π . Proof
Because Σ π ⊆ S . ⋄ Comment T π ( S ) is an exhaustive search. § The routine is a trial over all the solutions, that is, R π = T π (Σ π ). Proof
By Theorem § T π (Σ π ) = Σ π ∩ Σ π = Σ π = R π . ⋄ Comment T π ( R π ) = T π (Σ π ) = Σ π = R π . § § If A is an analogy, and π = x ? P π ( x ) is a problem, then Aπ isanother problem Aπ = x ? P Aπ ( x ). That is, A : P → P . Comment
So analogies transform a condition into a condition, P π into P Aπ in thisexample. Comment
Taking advantage of problem decomposition, see § Aπ , can be a composition of problems that are easier to resolve than theoriginal problem, π , see § § If Σ π = Σ Aπ , then we say that the analogy is conservative. Comment
If an analogy is not conservative, then a function T A to translate Σ Aπ to Σ π is required, because otherwise the analogy would be useless. § We will call function T A the translating function of analogy A . T A : 2 S → S and T A (Σ Aπ ) = Σ π . § An analogy followed by another one is an analogy.
Proof
Because any analogy transforms a problem into a problem: P → P . ⋄ Corollary
Analogies can be chained. ww.ramoncasares.com
PT 13 § Using only analogies we cannot resolve any problem.
Proof
Because using analogies we only get problems. ⋄ Comment
While routines R and trials T ( S ) are functions that return a set, P → S ,analogies A are functions that return a function, P → P . § We will write A ◦ T to express the composition of functions, where A is applied first and then T . Comment [ A ◦ T ]( x ) = T ( A ( x )). Diagram: x A −−→ A ( x ) T −−→ T ( A ( x )). Comment If A and A are analogies, then A ◦ A is also an analogy, by Lemma § § To resolve a problem by analogy A is to compose A ◦ ℜ ◦ T A , where ℜ is any resolution, and T A is the translating function of A . Diagrams: π A −−→ Aπ ℜ −−→ Σ Aπ T A −−→ Σ π or P A −−→ P ℜ −−→ S T A −−→ S . Comment
Analogy A is a translation from some original problem domain to someanalogue problem domain. Then, by Lemma § ℜ to resolvethe analogue problem. And, finally, we need to translate the solutions back to theoriginal domain. § The translating function of the composition A ◦ A ′ is T A ′ ◦ T A . Proof If ℜ = A ′ ◦ ℜ ′ ◦ T A ′ then we get A ◦ ( A ′ ◦ ℜ ′ ◦ T A ′ ) ◦ T A = A ◦ A ′ ◦ ℜ ′ ◦ T A ′ ◦ T A =( A ◦ A ′ ) ◦ ℜ ′ ◦ ( T A ′ ◦ T A ), because function composition is associative. Diagram: P A −−→ P A ′ −−→ P ℜ ′ −−→ S T A ′ −−−→ S | {z } ℜ T A −−→ S . ⋄ Corollary
The translating function of the composition A ◦ A . . . ◦ A n is T A n ◦ . . . ◦T A ◦ T A . That is: T A ◦ A ... ◦ A n = T A n ◦ . . . ◦ T A ◦ T A . Comment
This is how analogies can be chained. § The identity function, written I , transforms anything into itself: ∀ x, I ( x ) = x . Comment
The identity function I is an effectively calculable function, see § λ -definable; in λ -calculus, I = ( λx.x ). Comment
Identity I transforms π into π , I ( π ) = π , and P π into P π , I ( P π ) = P π . Comment
Identity I can work as an analogy: Iπ = I ( π ) = π . § The translating function of the identity analogy is the identity function: T I = I . Proof
Because I (Σ π ) = Σ π . Diagram: π I −→ π ℜ −−→ Σ π I −→ Σ π . ⋄ Comment
The identity analogy is conservative, see § § The identity I followed by any function f , or any function f followedby identity I , is equal to the function: ∀ f, I ◦ f = f = f ◦ I . Proof ∀ f, ∀ x, [ I ◦ f ]( x ) = f ( I ( x )) = f ( x ) = I ( f ( x )) = [ f ◦ I ]( x ). ⋄ Comment I ◦ ℜ ( Iπ ) ◦ T I = I ◦ ℜ ( π ) ◦ I = ℜ ( π ). ww.ramoncasares.com PT 14 § A ◦ T Aπ ( S ) ◦ T A , where A is an analogy, T Aπ ( S ) is a trial, and T A isthe translating function of A , is the general form of a resolution. Proof
If the analogy is the identity I , then the general form is reduced to T π ( S ),because T I = I , Iπ = π , so I ◦ T Iπ ( S ) ◦ I = T π ( S ), which is a trial. By Theorem § R π = T π ( R π ), so I ◦ T π ( R π ) ◦ I = T π ( R π ) = R π reducesthe general form to the routine. Resolving by analogy is, by definition, A ◦ ℜ ◦ T A ,and analogies can be chained, by Lemma § A ◦ A ◦ . . . ◦ A n = A , and by Lemma § T A = T A ◦ A ◦ ... ◦ A n = T A n ◦ . . . ◦ T A ◦ T A .Then A ◦ A ◦ . . . ◦ A n ◦ T Aπ ( S ) ◦ T A n ◦ . . . ◦ T A ◦ T A = A ◦ T Aπ ( S ) ◦ T A . ⋄ Summary
There are three ways to resolve a problem: routine R π = I ◦ T π ( R π ) ◦ I , trial T π ( S ) = I ◦ T π ( S ) ◦ I , and analogy A ◦ . . . ◦ A n ◦ T Aπ ( S ) ◦T A n ◦ . . . ◦T A = A ◦ T Aπ ( S ) ◦T A . § § A resolution ℜ : P → S is a valid resolution for a problem π ifit finds all the solutions of problem π and then halts. In other words, ℜ is a validresolution for π if it satisfies two conditions: that ℜ ( π ) is effectively calculable, andthat ℜ fits problem π , that is, that ℜ ( π ) = Σ π . Comment
If Σ π is infinite, | Σ π | ≥ ℵ , then a valid ℜ ( π ) does not halt, but it keepsbuilding Σ π forever. § A problem π is resolved if a valid resolution for π is known. Comment
To solve a problem we have to find one solution, see § § Once a problem is resolved, we can thereafter resolve it by routine.
Proof
Once a problem is resolved, we know all of its solutions, Σ π , and knowing Σ π ,we know its routine resolution, because R π = Σ π , see § ⋄ Proposition If π ∧ ρ is solvable, then by resolving π ∧ ρ both π and ρ are solved. § A problem is resolvable if there is a valid resolution for the problem,see § ℜ such that ℜ ( π ) is effectively calculable,and ℜ ( π ) = Σ π . Otherwise, the problem is unresolvable. Comment
A problem is resolvable if it can be resolved.
Comment
Resolved implies resolvable, but not the converse: Resolved ⇒ Resolvable. § For any Boolean-valued function P : S → B , we define the functionˇ P : B → S , called the inverse of condition P , as follows:ˇ P ( ⊤ ) = { x | [ P ( x ) = ⊤ ] } , ˇ P ( ⊥ ) = { x | [ P ( x ) = ⊥ ] } . § If P π ( x ) is the condition of a problem π , then ˇ P π ( ⊤ ) = Σ π andˇ P π ( ⊥ ) = Σ π = Σ ¯ π . Proof
Because ˇ P π ( ⊤ ) = { x | [ P π ( x ) = ⊤ ] } = { x | P π ( x ) } = Σ π , andˇ P π ( ⊥ ) = { x | [ P π ( x ) = ⊥ ] } = { x | ¬ P π ( x ) } = Σ π = Σ ¯ π , by Lemma § ⋄ ww.ramoncasares.com PT 15 § The inverse of the condition of a problem, provided it is an effectivelycalculable function, resolves the problem and its complementary by routine.
Proof By § § P π ( ⊤ ) = Σ π = R π , then ˇ P π ( ⊤ ) is the routine resolution of π , if ˇ P π ( ⊤ ) is effectively calculable, see § P π ( ⊥ ) is effectively calculable,then it resolves the complementary problem by routine, ˇ P π ( ⊥ ) = Σ ¯ π = R ¯ π . ⋄ Comment
It is a nice theorem, but how can we find the inverse of a condition? § The metaproblem of a problem, written Π π , is the problem offinding the valid resolutions for problem π . In other words, if π = x ? P π ( x ), thenΠ π = ℜ ? [ ℜ ( π ) = Σ π ]. Comment
The solutions of the metaproblems are the resolutions, Π S = R . Comment
The condition of the metaproblem, P Π π , is [ ℜ ( π ) = Σ π ], that is, P Π π ( ℜ ) =[ ℜ ( π ) = Σ π ], or using an α -conversion, P Π π ( x ) = [ x ( π ) = Σ π ]. § A metaproblem is a problem, that is, Π P ⊂ P . Proof
Because Π π = x ? P Π π ( x ), but some problems are not metaproblems. ⋄ Comment
A metaproblem is a problem because it has its two ingredients: there areseveral ways to resolve a problem, so there is freedom, but only the valid resolutionsresolve the problem, so there is a condition. § The metacondition P Π is P Π ( p, r ) = [ r ( p ) = Σ p ], for any problem p ∈ P , and for any resolution r ∈ R . Comment
Using another α -conversion, P Π ( π, x ) = [ x ( π ) = Σ π ] = P Π π ( x ). Comment Π π = x ? P Π ( π, x ). § Metaresolving is resolving the metaproblem to resolve the prob-lem.
Comment
Metaresolving is a kind of analogy. Diagram: π Π −−→ Π π Π ℜ −−−→ Σ Π π = { ℜ | [ ℜ ( π ) = Σ π ] } f c −−→ ℜ c ( π ) −−→ ℜ c ( π ) = Σ π . Function f c is a choice function, and the last calculation, noted ( π ), means to apply π as the argument, not as the function. If you only metasolve, then you don’t needto choose. In any case, the translating function of metaresolving is T Π = f c ◦ ( π ).Then we can draw the following diagrams: π Π −−→ Π π Π ℜ −−−→ Σ Π π T Π −−→ Σ π or P Π −−→ Π P Π ℜ −−−→ Π S = 2 R T Π −−→ S . § The metaproblem Π π of some problem π is solvable if, and only if, theproblem π is resolvable, that is, Π π is solvable ⇔ π is resolvable. Proof
If Π π is solvable, then there is a solution to it, see § π , see § π is resolvable, see § π isresolvable, then there is a valid resolution for it, see § π , see § π is solvable, see § ⋄ Corollary
To solve the metaproblem Π π of problem π is to resolve problem π . Proof
Because to resolve problem π is to find a valid resolution for π , see § π is to find a solution to Π π , see § π , see § ⋄ Comment
And again, R = Π S . ww.ramoncasares.com PT 16 § The set of the valid resolutions for problem π is the routine resolutionof its metaproblem Π π , that is, { ℜ | [ ℜ ( π ) = Σ π ] } = R Π π . Proof R Π π = Σ Π π , by the definition of routine, see § Π π = { ℜ | [ ℜ ( π ) = Σ π ] } , by the definition of Π π , see § ⋄ § § The meta n -metaproblem of π , Π n Π π , is (the metaproblem of) n themetaproblem of π , where n ∈ N . Special case
The meta-metaproblem of π , ΠΠ π = Π Π π , is the metaproblem of themetaproblem of π . Examples Π Π π = Π π . Π Π π = ΠΠ π = Π π . Π Π π = ΠΠΠ π = Π π . Comment
From Π S = R , we get ΠΠ S = Π R and Π n Π S = Π n R . Comment
The condition of the meta n -metaproblem of π , P Π n Π π , where n ∈ N , is: P Π n Π π ( x ) = [ x (Π n π ) = Σ Π n π ]. Examples P Π Π π ( x ) = [ x (Π π ) = Σ Π π ] = [ x ( π ) = Σ π ] = P Π π ( x ). P Π Π π ( x ) = [ x (Π π ) = Σ Π π ] = [ x (Π π ) = Σ Π π ] = P ΠΠ π ( x ). § A meta n -metaproblem is a problem, where n ∈ N . Proof If n >
0, then Π n Π π = x ? P Π n Π π ( x ). For n = 0, see § ⋄ Corollary S n ∈ N Π n Π P ⊂ P . § The meta n -metacondition P Π n Π , with n ∈ N , p ∈ P , and r ∈ R is: P Π n Π ( p, r ) = [ r (Π n p ) = Σ Π n p ]. Comment
Using an α -conversion, P Π n Π ( π, x ) = [ x (Π n π ) = Σ Π n π ] = P Π n Π π ( x ). Example P Π Π ( π, x ) = P ΠΠ ( π, x ) = [ x (Π π ) = Σ Π π ] = [ x (Π π ) = Σ Π π ]. § P Π n Π ( π, x ) = P Π (Π n π, x ), where n ∈ N . Proof By § P Π (Π n π, x ) = [ x (Π n π ) = Σ Π n π ] = P Π n Π ( π, x ). ⋄ Special Case P ΠΠ ( π, x ) = P Π (Π π, x ). Comment
The meta-metacondition is the metacondition of the metaproblem. § A meta n -metaproblem is a metaproblem, where n ∈ N . Proof If n >
0, Π n Π π = x ? P Π (Π n π, x ), and Π Π π = Π π = x ? P Π ( π, x ). ⋄ Corollary S n ∈ N Π n Π P = Π P . § We have the following infinite series of mathematical objects: S , P = 2 S , R = Π S = 2 S → S , Π P = 2 S → S , Π R = ΠΠ S = 2 S → S → S → S , . . . Proof P = S → B = 2 S , by Theorems § § S = R , by the metaproblem definition, see § R = P → S = 2 S → S .Π P = Π S → B = R → B = 2 R = 2 S → S .Π R = Π P → Π S = 2 R → R = 2 S → S → S → S .And so on. ⋄§ There is only one level of problem meta-ness.
Proof
By Lemma § n -metaproblem is a metaproblem, andevery metaproblem is a meta -metaproblem, so S n ∈ N Π n Π P = Π P ⊂ P . ⋄ Comment
While a problem condition is any predicate, P ( x ), a metaproblem conditionis a specific kind of predicate, namely, P Π ( p, r ) = [ r ( p ) = Σ p ]. And any meta n -metaproblem condition, P Π n Π , is the same specific predicate P Π , see § ww.ramoncasares.com PT 17
Comment
We are assuming that functions are free to take functions as arguments. Seethat, in predicate P Π ( p, r ) = [ r ( p ) = Σ p ], argument r is a function in Π n R that takes p ∈ Π n P as argument. Therefore, the theorem holds unconditionally for λ -definablefunctions, including predicates, see § § § § § There are five types of resolution.
Proof
From Theorem § R π , T π ( S ), and A ◦ T Aπ ( S ) ◦ T A . This shows that there are several ways of resolving, sochoosing a resolution that find solutions to the original problem π is another problem,the metaproblem Π π , see § § T Π π ( R ), and A ◦ T A Π π ( R ) ◦ T A . Finally, by § n -metaproblems. ⋄ Comment
We will call them: routine R π , trial T π ( S ), analogy A ◦ T Aπ ( S ) ◦ T A , meta-trial T Π π ( R ), and meta-analogy A ◦ T A Π π ( R ) ◦ T A . The first three can also be calledmeta-routines. § The diagram for the meta-trial, or trial of the metaproblem, is: π Π −−→ Π π T Π π ( R ) −−−−−→ Σ Π π T Π −−→ Σ π . And the diagram for the meta-analogy, or analogy of the metaproblem, is: π Π −−→ Π π A −−→ A Π π T A Π π ( R ) −−−−−−→ Σ A Π π T A −−→ Σ Π π T Π −−→ Σ π . See that A : Π P → Π P = 2 S → S → S → S and T A : 2 Π S → Π S = 2 S → S → S → S ,using § § § § A computation is any manipulation of a string of symbols, irrespec-tive of the symbols meanings, but according to a finite set of well-defined rules.
Comment
Computing is any mechanical transformation of a string of symbols. § A computing device, or computer, is any mechanism that can per-form computations.
Comment
The prototype of computing device is a Turing machine, see Turing (1936). ww.ramoncasares.com
PT 18 § The Turing machine has two parts: the processor P , which is a finitestate automaton, and an infinite tape, h i , which in any moment contains only a finitenumber of symbols. Comment
In the case of a processor of a Turing machine, the output alphabet O , thatis, the finite set of output symbols, has to be: O = I + × { l, h, r } , where I is the finitenot empty input alphabet, I + = I ∪ { b } , where b / ∈ I is blank , and l , h , and r mean left , halt , and right . Then its transition function is T : S × I + → S × I + × { l, h, r } ,where S is the finite set of internal states. And the strings that the Turing machinetransforms are sequences of symbols taken from set I . § We will refer to the set of Turing machines as T . We will refer tothe set of the strings of symbols as E . Comment
Because all Turing machines tapes are equal, the processor defines the Tur-ing machine, and therefore we will refer to the Turing machine with processor P asthe Turing machine P , and then P ∈ T . We will refer to the string of symbols writtenon the tape as the expression e ∈ E . § The set of expressions is countable, that is, | E | = | N | = ℵ . Proof
Let I be any finite alphabet, and s its cardinality, that is, s is the number ofsymbols, s = | I | >
0. We write I n the set of strings of length n , so | I n | = s n . Then E = S n ∈ N I n , and we can define a bijection between E and N this way: it maps theempty string in I to 0, it maps the s strings in I to the next s numbers, it maps the s strings in I to the next s numbers, and so on. Note that ordering the symbolsin I , we can order alphabetically the strings in each I n . ⋄ Comment
Most real numbers are not expressible. See Turing (1936) §
10 for details;but, for example, transcendental numbers π and e are computable, page 256. § We will use the notation Ph e i ֒ → r to indicate that, if we write theexpression e ∈ E on the tape of the Turing machine with processor P and we leaveit running, then when it halts we will find the expression r ∈ E on the tape. If, onthe contrary, the Turing machine P does not halt when we write the expression w ,then we would say that w is a paradox in P , and we would indicate this as follows: Ph w i ֒ → ∞ . § E + = E ∪ {∞} . Comment
Some computations do not halt, so we need ∞ to refer to them. Note that ∞ / ∈ E , but ∞ ∈ E + . So E ⊂ E + . § For each Turing machine
P ∈ T we define a function F P : E → E + ,this way: F P ( e ) = (cid:26) r if Ph e i ֒ → r ∞ if Ph e i ֒ → ∞ . Comment If ∀ e ∈ E , F P ( e ) = F Q ( e ), then we say that Turing machines P and Q arebehaviorally equivalent, P ≡ F Q , or that P and Q implement the same function. § We say that a function is computable if there is a Turing machinethat implements the function. ww.ramoncasares.com
PT 19 § For each Turing machine we can define a unique finite string of symbols,that is, ∃ c : T → E such that P = Q ⇔ c ( P ) = c ( Q ). Proof
Proved by Turing (1936), §
5. Turing machines are defined by their processors,which are finite state automata. And every finite state automaton is defined bythe table that describes its transition function T in full, which is a finite table ofexpressions referring to internal states, input symbols, and output symbols. A tablecan be converted to a string just using an additional symbol for the end of line, andanother symbol for the end of cell. To assure uniqueness, we have to impose someorder on the lines and on the cells. ⋄ Comment c ( P ) ∈ E is the string of symbols that represents the Turing machine P ∈ T . § We will refer to p = c ( P ) as a program, and to the set of programsas P . The set of programs is a proper subset of the set of expressions, P ⊂ E . § The program isomorphism is the natural isomorphism that relateseach Turing machine
P ∈ T with the expression describing it, c ( P ) = p ∈ P . Thatis, T ⇔ P : P ↔ c ( P ). Comment
Now, T ∼ = P ⊂ E . § The set of Turing machines is countable, that is, | T | = | N | = ℵ . Proof
Proved by Turing (1936), §
5. Using the program isomorphism, see § p = c ( P ). We canorder the programs, because they are finite strings of symbols, for example first bylength, and then those of a given length by some kind of alphabetical order. Onceordered, we can assign a natural number to each one. ⋄§ All computing sets are countable, that is, | T | = | E | = ℵ . Proof
By Lemmas § § ⋄ Comment
All computing is about countable sets. Computing is counting. § § There is a Turing machine, called universal Turing machine, U , thatcan compute anything that any Turing machine can compute. That is: ∃ U ∈ T | ∀P ∈ T , ∀ d ∈ E , U h c ( P ) d i = Ph d i . Proof
Proved by Turing (1936), § § ⋄ Comment
The equality means that if Ph d i ֒ → r then U h c ( P ) d i ֒ → r , and the converse,and also that if Ph d i ֒ → ∞ then U h c ( P ) d i ֒ → ∞ , and the converse. That is, U h c ( P ) i ≡ F P . To complete the definition, if e / ∈ P , then U h e d i ֒ → e d . § We will refer to the set of universal Turing machines as U . Comment
The set of universal Turing machines is a proper subset of the set of Turingmachines, U ⊂ T . § For each universal Turing machine U there is a universal program u . Proof
Universal Turing machines are Turing machines, and u = c ( U ). Then, by theprogram isomorphism, see § u = U . ⋄ Comment
Given u = c ( U ) and p = c ( P ), then U h p d i = Ph d i and U h u p d i = U h p d i ,so u is the identity for programs, and U h u u p d i = U h u p d i = U h p d i = Ph d i . ww.ramoncasares.com PT 20 § The terminating condition P σ : T → B is: P σ ( P ) = (cid:26) ⊥ if ∃ w ∈ E , Ph w i ֒ → ∞⊤ otherwise . Comment
A terminating Turing machine always halts. There are not paradoxes in aterminating Turing machine. While Turing machines implement partial functions, E → E + , see § E → E . § The terminating problem is σ = p ? P σ ( p ).The non-terminating problem is ¯ σ = p ? ¬ P σ ( p ). Comment
The terminating problem follows from the condition isomorphism of prob-lems, see § P σ .The non-terminating problem is derived from the terminating one by negation, see § Comment
The set of terminating Turing machines is Σ σ ,and the set of non-terminating Turing machines is Σ ¯ σ . Proposition Σ σ and Σ ¯ σ are a partition of T , because Σ σ ∩ Σ ¯ σ = ∅ and Σ σ ∪ Σ ¯ σ = T . § We will call a = c ( P σ ) ∈ E , where P σ ∈ Σ σ , an algorithm. Comment ∀ d , P σ h d i ֒ → r = ∞ ⇔ ∀ d , U h a d i ֒ → r = ∞ . Comment
An algorithm is the expression of a computation that always halts.
Notation
We will refer to the set of algorithms as A . Comment A ⊂ P ⊂ E . § Universal Turing machines are non-terminating, that is, U ⊂ Σ ¯ σ ⊂ T . Proof
Because there are paradoxes in some Turing machines. For example, for Turingmachine W , that has not any h ( halt ) in its transition table, every expression is aparadox. That is, ∃P ∈ T , ∃ w ∈ E , Ph w i ֒ → ∞ ⇒ ∀U ∈ U , U h c ( P ) w i ֒ → ∞ . ⋄ Comment
If expression w is a paradox in P , then expression c ( P ) w is a paradox in U . Then, U ∈ Σ ¯ σ . § A computing device is Turing complete if it can compute what-ever any Turing machine can compute. We will call every Turing complete device auniversal computer.
Comment
The prototype of universal computer is a universal Turing machine, U . Comment
The Turing machine, as it was presented by Turing (1936), models thecalculations done by a person. This means that we can compute whatever any Turingmachine can compute provided we have enough time and memory, and therefore weare Turing complete provided we have enough time and memory. § All universal computers are equivalent.
Proof
G¨odel and Herbrand recursiveness, Church λ -definability, and Turing com-putability are equivalent, because Kleene (1936) showed that every recursive functionis λ -definable, and the converse, and then Turing (1937) showed that every λ -definablefunction is computable, and that every computable function is recursive. ⋄ Comment
A universal Turing machine is equivalent to a λ -calculus interpreter, wherea λ -calculus interpreter is a device that can perform any λ -calculus reduction. Auniversal Turing machine is equivalent to a mathematician calculating formally, andwithout errors, any recursive function. ww.ramoncasares.com PT 21
Comment
The universal Turing machine, the λ -calculus interpreter, and the mathe-matician, who is a person, are equal in computing power. And all of them are Turingcomplete. § Whenever we apply a general statement to a finite universal comput-ing device, we should add a cautious ‘provided it has enough time and memory’.
Comment
Although the finite universal computer can perform each and every step ofthe computation exactly the same as the unrestricted universal computer, the finiteuniversal computer could meet some limitations of time or memory that would preventit to complete the computation. In that case, the same finite universal computer,provided with some additional time and some more memory, would perform somemore computing steps exactly the same as the unrestricted universal computer. Thisextension procedure can be repeated as desired to close the gap between the finiteand the unrestricted universal computer.
Comment
We will understand that the proviso ‘provided it has enough time and mem-ory’ is implicitly stated whenever we refer to a finite universal computing device. § Because all universal computers are equivalent, we can use anyof them, let us call the one used U , and then drop every U from the formulas, and justexamine expressions, that is, elements in E . In case we need to note a non-haltingcomputation, we will use ∞ . Comment
Using the convention is as if we were always looking inside the tape of U . Given a universal computer, U , computing is about expressions manipulatingexpressions. Example
Formula
U h c ( P ) d i ֒ → r is reduced to h c ( P ) d i ֒ → r , and even to h p d i ֒ → r ,using the rewriting rule: ∀P ∈ T , c ( P ) = p . If the universal computer is a λ -calculusinterpreter, then this is usually written as the β -reduction ( p d ) → r , where the lefthand side is a λ -application, and p is defined by some λ -abstraction. § For each program p ∈ P we define a function F p : E → E + , thisway: F p ( e ) = (cid:26) r if h p e i ֒ → r ∞ if h p e i ֒ → ∞ . Comment If ∀ e ∈ E , F p ( e ) = F q ( e ), then we say that programs p and q are behaviorallyequivalent, p ≡ F q , or that p and q implement the same function. § ∀P ∈ T , F P = F p , where p = c ( P ). Proof ∀ d ∈ E , ∀P ∈ T , F p ( d ) = F P ( d ), see § U h c ( P ) d i = Ph d i , byTheorem § F P = F p when the universal computer is a universalTuring machine, U . Theorem § ⋄ Comment P and p implement the same function. Comment
This theorem is a consequence of the program isomorphism, see § T ∼ = P implies that ≡ F ↔ ≡ F , so P ≡ F Q ⇔ p ≡ F q . Corollary F U = F u , where u = c ( U ). ww.ramoncasares.com PT 22 § § What is effectively calculable is computable.
Comment
This is Church’s thesis, or rather Turing’s thesis, as it was expressed byGandy (1980). There, ‘something is effectively calculable’ if its results can be foundby some purely mechanical process, see § ∗ F ⊆ T . Comment ‘What is computable is effectively calculable’, or T ⊆ ∗ F , is the converse ofTuring’s thesis. And it is obvious that if a Turing machine can compute a function,then the function is effectively calculable, see § ∗ F = T , and | ∗ F | = ℵ , by § § An effectively calculable function is not an input to output mapping;it is a process to calculate the output from the input.
Example
To multiply a number expressed in binary by two we can append a ‘0’ to it,which is an effectively calculable function that we will call app0 . But the completememoization of the same function, which we will call memoby2 , is not effectivelycalculable because it would require an infinite quantity of memory. And therefore, app0 = memoby2 . § We will call every universe where the Turing’s thesis is true a Turinguniverse. When we want to note that something is true in a Turing universe, we willuse an asterisk, so A ∗ = B means that A = B if the Turing’s thesis stands. Examples ∗ F ∗ = T and | ∗ F | ∗ = ℵ . Comment
The Turing’s thesis affirms that this is a Turing universe. In any Turinguniverse the Turing’s thesis is a law of nature, as it was defended by Post (1936), lastparagraph. Then a Turing universe can also be called a Post universe.
Comment
While the Turing’s thesis is true, you can ignore the asterisks. § Universal computers are* the most capable computing devices.
Proof
If Turing’s thesis stands, see § § § ⋄§ There are definable functions that no Turing machine can compute.
Proof
You can use a diagonal argument, or work from other theorems that use thediagonal argument. For example, the set of Turing machines is countable, see § | T | = | N | = ℵ , while the possible number of predicates on natural numbers, that is,the number of functions N → B , is 2 | N | = 2 ℵ , which is not countable, | T | = | N | = ℵ < ℵ = 2 | N | . This uses Cantor’s theorem, | S | < | S | , with its diagonal argument.So there are not enough Turing machines to compute every definable function. ⋄ Corollary
Universal computers cannot compute every definable function.
Comment
If the Turing’s thesis stands, see § § Comment
There are* more mappings than processes. § The identity Turing machine, I , just halts. Comment
It does nearly nothing. But, wait! ww.ramoncasares.com
PT 23 § ∀ x ∈ E , Ih x i ֒ → x , where I is the identity Turing machine. Proof
Whatever expression x ∈ E is written on the tape of I , that very same expression x is written when I halts, because halting is all what I does. ⋄ Comment I does not touch the expression. § The identity Turing machine is terminating, that is,
I ∈ Σ σ . Proof
The identity Turing machine, which just halts, is terminating, see § ⋄ Comment I behaves, because sometimes ‘you can look, but you better not touch’. § The identity Turing machine I : E → E is* the identity function i : S → S such that ∀ x ∈ S , i ( x ) = x , that is, I ∗ = i . Proof
The identity function i is an effectively calculable function, see § § J suchthat ∀ x ∈ E , J h x i ֒ → x . By Lemma § J is the identityTuring machine I . ⋄ Comment If i = c ( I ), then U h i p d i = Ih p d i ֒ → p d , and U h u p d i = U h p d i = Ph d i ֒ → r ,or ∞ , see § i is the literal identity for expressions, or quotation, and u isthe functional identity for programs, or evaluation. Both are computable, but I ∈ Σ σ and U ∈ Σ ¯ σ , see § § Everything is* an expression, that is, E ∗ = S . Proof S is the set of everything, see § T , and expressions, E , see § § T ⊂ E . Therefore, in computingeverything is an expression. And now, if the Turing’s thesis stands, see § § ∀ x ∈ S , x = i ( x ) ∗ = Ih x i ֒ → x ∈ E . The converse, ∀ x ∈ E , x ← ֓ Ih x i = i ( x ) = x ∈ S , holds irrespective of Turing’s thesis. Therefore, S ∗ = E . ⋄ Comment
We will write x to indicate a computing point of view of x , but ∀ x, x ∗ = x .For example, i ∗ = i . § The set of solutions S is* countable, that is, | S | ∗ = ℵ . Proof S ∗ = E , by § | E | = | N | = ℵ , by § | S | ∗ = | E | = ℵ . ⋄ Comment
We will refer to the set of solutions in a Turing universe as S ∗ . So we canalso write this lemma as | S ∗ | = ℵ . § Resolving is* computing, that is, T ∗ = R . Proof
From Theorem § P , or its equivalent program p , or a λ -function of the λ -calculus, is* equivalent to whatever transforms sets in set theory, that is, an effectivelycalculable function, and it is* also equivalent to whatever transforms problems inproblem theory, that is, a resolution ℜ . Therefore, resolving is* computing, R ∗ = T . ⋄ Comment ∗ f ∗ ≡ P ≡ p ∗ ≡ ℜ , and ∗ F ∗ = T ∗ = R . Comment
We can define functions that are not effectively calculable, see § ww.ramoncasares.com PT 24
Corollary
Metasolutions are* effectively calculable functions, that is, Π S ∗ = ∗ F . Proof
Because R = Π S , see § S = R ∗ = ∗ F . ⋄§ The set of resolutions R is* countable, that is, | R | ∗ = ℵ . Proof R ∗ = T , by § | T | = | N | = ℵ , by § | R | ∗ = | T | = ℵ . ⋄ Comment
We will refer to the set of resolutions in a Turing universe as R ∗ . So we canalso write this lemma as | R ∗ | = ℵ . § Predicate P δ s , where P δ s ( x ) = [ x = s ], is* effectively calculable. Proof
Both s and x are* expressions, by § s = s s . . . s n , and x = x x . . . x m . Then we can define a Turing machine with n + 2states, that starts in state 1, and that when some string x is written on its tape, itscans the string x , symbol by symbol, from the leftest one, this way: 1) in state i ,with 1 ≤ i ≤ n , if the read symbol is s i , then it writes a blank , goes to state i + 1,and moves to the right , but if the read symbol is not s i , then it writes a blank , goesto state 0, and moves to the right ; 2) in state n + 1, if the read symbol is blank , thenit writes a ⊤ , goes to state 0, and halt s, but if the read symbol is not blank , thenit writes a blank , goes to state 0, and moves to the right ; 3) in state 0, if the readsymbol is not blank , then it writes a blank , goes to state 0, and moves to the right ,but if the read symbol is blank , then it writes a ⊥ , goes to state 0, and halt s. ThisTuring machine implements P δ s , and therefore P δ s is computable. ⋄ Corollary
Problem δ s = x ? [ x = s ] is* expressible. Proof
Because problem δ s condition P δ s is* effectively calculable, see § ⋄ Comment
Problem δ s is used in the proof of Theorem § Corollary
The only solution to problem δ s is s , so Σ δ s = { s } ∈ S . Proof
Because Σ δ s = { x | P δ s ( x ) } = { x | [ x = s ] } = { s } . ⋄§ The set of problems P is* countable, that is, | P | ∗ = ℵ . Comment
We will refer to the set of problems in a Turing universe as P ∗ . So we canalso write this lemma as | P ∗ | = ℵ . If the condition of a problem is computable, thenthe problem is in P ∗ ; δ s ∈ P ∗ , for example. Proof
Problem δ s is* expressible, see § δ S ∗ = { δ s | s ∈ S ∗ } ⊆ P ∗ becauseeach δ s ∈ P ∗ , and | δ S ∗ | = | S ∗ | because there is a bijection δ S ∗ ⇔ S ∗ : δ s ↔ s . Also,by Theorem § P ∗ ⊆ E . Therefore, δ S ∗ ⊆ P ∗ ⊆ E , and | δ S ∗ | = | S ∗ | = ℵ = | E | ,and then, by the Cantor-Bernstein-Schr¨oder theorem, | P ∗ | = ℵ . ⋄ Comment
The Cantor-Bernstein-Schr¨oder theorem is Theorem B of §
2, page 484, inCantor (1895). We have really used the equivalent Theorem C, in the same page. § All problem sets are* countable, that is, | S ∗ | = | P ∗ | = | R ∗ | = ℵ . Proof
By Lemmas § § § ⋄ § § A full resolution machine is a device that can execute any resolution. § A full resolution machine is* a Turing complete device.
Proof
By Theorem § ℜ ∗ ≡ P . This means that toachieve the maximum resolving power is* to achieve the maximum computing power,which is* the computing power of a universal computer, by Theorem § ⋄ ww.ramoncasares.com PT 25
Comment
To execute any resolution ℜ : P → S , the full resolution machine has tocalculate functions that can take functions and that can return functions withoutlimitations, as 2 S → S → S → S for meta-analogies, see § λ -calculusinterpreter, or an equivalent computing device, for example U . Comment
This means that problem resolving is* equal to computing, and then fullproblem resolving is* equal to universal computing.
Corollary
A full resolution machine* is a universal computer.
Proof
Because a full resolution machine is* a universal computer. ⋄ Comment
Now we will state two equivalences between computing theory and problemtheory concepts that are true in any Turing universe, and that are needed to showthe limitations of full resolution machines. § A set is recursively enumerable if there is a Turing machine thatgenerates all of its members, and then halts.
Comment
If the set is infinite, the Turing machine will keep generating its membersforever.
Definition
A set is computable if it is recursively enumerable. § Resolvable in problem theory is* equivalent to recursively enumerablein computing theory, that is,Resolvable ∗ = Recursively Enumerable . Proof
To see that a problem is resolvable if, and only if, the set of its solutions isrecursively enumerable, just compare the definition of resolvable problem, in § § ℜ of the resolvable problem to the Turing machine of therecursively enumerable set, a gap that we can bridge with the help of Theorem § § π or to the set of its solutions Σ π . Then we can say that a problem isrecursively enumerable, or that a set is resolvable. ⋄§ A set is recursive if its characteristic function can be computed bya Turing machine that always halts. § Expressible in problem theory is* equivalent to recursive in comput-ing theory, that is, Expressible ∗ = Recursive . Proof
The condition of a problem, P π , is the characteristic function of the set of itssolutions, because Σ π = { s | P π ( s ) } , see § § P π can be computed bya Turing machine that always halts. So the condition P π is an effectively calculablefunction, and therefore the problem is expressible, see § § § π or to the set of its solutions Σ π . Then, we can say that a problemis recursive, or that a set is expressible. ⋄ ww.ramoncasares.com PT 26 § The limitations of full resolution machines are* the limitations of uni-versal computers.
Proof
Because a full resolution machine is* a universal computer, see § ⋄ Comment
Even if universal computers are the most capable computers, they cannotcompute everything, see § § A full resolution machine can* execute any resolution, but it cannot*express some problems.
Proof
There is a recursively enumerable set that is not recursive; this is the lasttheorem in Post (1944) §
1. Translating, by Theorems § § ⋄ Comment
This is the problem limit of full resolution machines*.
Comment
That last theorem in Post (1944) §
1, page 291, is an abstract form of G¨odel’sincompleteness theorem, see Post (1944) § § A full resolution machine can* execute any resolution, but it cannot*resolve some problems.
Proof
Let us call κ some problem that is resolvable but not expressible, see § ∃ℜ | ℜ ( κ ) = Σ κ , but P κ | P κ ( x ) = [ x ∈ Σ κ ]. Note that | Σ κ | ≥ ℵ ,because otherwise ∃ P κ . Then its metaproblem Π κ is solvable but not resolvable. Π κ is solvable because κ is resolvable, see § ℜ is a solution toΠ κ . For Π κ to be resolvable there should be a resolution that would find ‘all thesolutions of Π κ ’, that is, ‘all the valid resolutions for κ ’. But, whenever a possiblevalid resolution for κ , let us call it ℜ ′ , generates a value not yet generated by ℜ , let uscall it z , we cannot decide whether z ∈ Σ κ and it will be eventually generated by ℜ ,or if z / ∈ Σ κ and it will never be generated by ℜ ; remember that κ is not expressible, P κ . And, not being able to decide on z , we cannot decide whether ℜ ′ is a validresolution for κ or not. ⋄ Comment
This is the resolution limit of full resolution machines*.
Comment
Problem κ is named after the complete set K of Post (1944), § § A full resolution machine can execute any resolution, but it cannotsolve some problems.
Proof
Simply because some problems have not any solution, Σ π = {} = ∅ . ⋄ Comment
This is the solution limit of full resolution machines, which also applies tofull resolution machines*.
Comment
An unsolvable problem can be resolved by showing that it has not anysolution. For example, the decision problem of the halting problem, ∆ η , see § § § A full resolution machine* can execute any resolution, but it can-not express some problems (problem limit), and it cannot resolve some problems(resolution limit), and it cannot solve some problems (solution limit).
Proof
By Lemmas § § § ⋄ Comment
Full resolution machines* have limitations on each of the three main con-cepts of the problem theory.Problem Resolution −−−−−−−−−−−−→ {
Solution } ww.ramoncasares.com PT 27 § § The decision problem of a problem π = x ? P π ( x ), written ∆ π , is:∆ π = P ? [ P ∈ Σ σ ] ∧ [ ∀ ( x ∗ = x ) , Ph x i ∗ = P π ( x )] . Comment
A solution to the decision problem ∆ π of some original problem π is a Turingmachine P that always halts and that computes the original problem condition P π for any input. Decision problems are only defined in Turing universes, where x ∗ = x by Theorem § Comment
This definition follows Post (1944), page 287. § The halting condition P η : T × E → B is: P η ( P , d ) = (cid:26) ⊥ if Ph d i ֒ → ∞⊤ otherwise . § The halting problem is η = ( p, d )? P η ( p, d ). Comment
The halting problem η corresponds to the halting condition P η by the con-dition isomorphism of problems, see § Comment P σ ( p ) = V d ∈ E P η ( p, d ), see § σ = V d ∈ E η , by § § § The decision problem of the halting problem, ∆ η , is:∆ η = H ? [ H ∈ Σ σ ] ∧ [ ∀P ∈ T , ∀ d ∈ E , Hh c ( P ) d i = P η ( P , d )] . § The decision problem of the halting problem ∆ η has not any solution. Proof
Turing (1936), §
8, resolved that ∆ η is unsolvable. ⋄ Comment
There is not any Turing machine that always halts and that compute P η foreach possible input. There is not any algorithm a ∈ A that would compute P η ( p , d )for every pair ( p , d ) ∈ P × E . § The decision problem ∆ π of some problem π is solvable if, and only if,the problem π is expressible*, that is, ∆ π is solvable ⇔ π is expressible*. Proof
From solvable to expressible. That the decision problem ∆ π is solvable, see § P π for each possible input. Therefore, P π is effectively calculable, see § π is expressible, see § π es expressible, thenits condition P π is an effectively calculable function, see § § P that can compute P π exactly as the effectively calculable function. P always halts,because P π is a condition, so its result is finite. Therefore, the decision problem ∆ π of the problem has a solution, P , and then ∆ π is solvable, see § ⋄ Corollary
The halting problem η is not expressible*. Proof
The decision problem of the halting problem, ∆ η , is not solvable, see § η is not expressible*. ⋄ Comment
The halting problem η is inexpressible*, but solvable. While the decisionproblem of the halting problem ∆ η is unsolvable, the halting problem η has manysolutions. ww.ramoncasares.com PT 28 § The following equivalences stand:∆ π is solvable ∗ ⇔ π is expressible , Π π is solvable ⇔ π is resolvable ,π is solvable ⇔ π is solvable . Proof
The last one is trivial, and the other two equivalences were already proved byLemmas § § ⋄§ A problem π can be: expressible* ( E ) or not expressible* ( E ),resolvable* ( R ) or not resolvable* ( R ), and solvable ( S ) or not solvable ( S ). Comment
An expressible problem is* equivalent to a recursive set, by Theorem § § Comment
Then R is the set of computable sets, see § Comment
Not every combination is possible. § If a problem is expressible*, then it is resolvable*, that is,
E ⊂ R . Proof
Because every recursive set is recursively enumerable,
E ⊆ R . This is a corollaryto the first theorem in Post (1944) §
1. And
E 6 = R , see the proof of Lemma § § § ⋄ Comment
The first theorem in Post (1944) §
1, page 290, states that a set M is recursiveif and only if both the set M and its complement M are recursively enumerable. § If a problem is not solvable, then it is expressible*, that is,
S ⊂ E . Proof
If a problem ν is not solvable, ν ∈ S , then Σ ν = {} , see § ν is acontradictory problem, see § P ν is the contradiction P ¯ τ , thatis, ∀ x, P ν ( x ) = P ¯ τ ( x ) = ⊥ . So P ν = P ¯ τ is an effectively calculable function, see § ν is expressible, see § S 6 = E ,because ( x ? [2 x = x ]) ∈ S ∩ E . ⋄ Comment
Being expressible*, by Lemma § ν is also resolvable*: S ⊂ E ⊂ R . § Regarding expressibility* E , resolvability* R , and solvability S , thetopology of the problem space is: S ⊂ E ⊂ R ⊂ P . Proof
By Lemmas § § E R S
Example & Comment ⊤ ⊤ ⊤ x ? [2 x = x ] ⊤ ⊤ ⊥ x ? [2 x = x ] ∧ [ x > ⊤ ⊥ ⊤ None, by Lemma § ⊤ ⊥ ⊥ None, by Lemma § ⊥ ⊤ ⊤ κ , see § ⊥ ⊤ ⊥ None, by Lemma § ⊥ ⊥ ⊤ Π κ , see § ⊥ ⊥ ⊥ None, by Lemma § ⋄ Corollary
Then, { S , E ∩ S , R ∩ E , R } is a partition of P . ww.ramoncasares.com PT 29
Comment
See below, in § E ∗ = P , and then P ∗ ⊂ P . Also R ∗ = 2 S . § We say that a problem is finite, if the set of its solutions is finite.We will refer to the set of finite problems as F . That is, F = { π | | Σ π | < ℵ } . § The set of finite problems F is a proper subset of the set of expressibleproblems E . The set of not solvable problems S is a proper subset of the set of finiteproblems F . That is, S ⊂ F ⊂ E . Proof
F ⊂ E because all finite sets are recursive, but not the converse.
S ⊂ F because ∀ ν ∈ S , | Σ ν | = 0 < ℵ , but ( x ? [2 x = x ]) ∈ S ∩ F . ⋄ Proposition
Including F , the topology of P is: S ⊂ F ⊂ E ⊂ R ⊂ P . Corollary
The topology
S ⊂ F ⊂ E ⊂ R ⊂ P partitions the problem space P intofive non-empty places: S , F ∩ S , E ∩ F , R ∩ E , and R . § The upper part of this topology is further refined by the so calledTuring degrees of unsolvability, that we will call Turing degrees of inexpressibility.Turing degree zero, , corresponds to the first three places, because E = . Comment
Then, |E | = | | = ℵ , |R| = ℵ , and | P | = 2 ℵ > ℵ . To complete thecardinalities, |S| = 1, so |S| = 2 ℵ , and |F | = ℵ . § Noting E p the set of problems defined by a condition that can becomputed in polynomial time, and R p the set of problems that can be resolved inpolynomial time, then E p ⊂ E and R p ⊂ R . The so called ‘P = NP?’ questionasks if E p = R p , because P = E p and NP = R p , and then it should be called the‘ E p = R p ?’ question. See that the general question ‘ E = R ?’ was answered negativelyby Lemma § E ⊂ R , and that E p ⊆ R p . Comment
A similar question is ‘ E p \ {∅} = S p ?’, where S p is the set of problems thatcan be solved in polynomial time, so S p ⊂ S . The corresponding general question isalso answered negatively, because P = S ∪ {∅} , so
E \ {∅} ⊂ S . § § § In this Section §
5, we will always be inside a Turing universe, see § § A resolver is a device that takes problems and returns solutions.
Comment
A resolver executes resolutions.
Comment
After Theorem § ℜ ∈ R to the computingdevice that executes the resolution P ∈ T , that is, ℜ = P . § We will call the domain of S semantics. We will call the domain of S → S syntax. Comment As λ -calculus shows, we only need functions to implement any syntax. Comment
By Theorem § S , including S → S . But this is bothmathematically impossible, by Cantor’s theorem, and practically not interesting. Example
Using a practical example, if the problem is the survival problem, so somebehaviors keep the resolver alive, and the rest cause the death of the resolver, then S ww.ramoncasares.com PT 30 is the set of behaviors, and it does not include anything that is not a behavior, noteven predicates on behaviors, nor functions. Note that the condition of the survivalproblem, which is satisfied if the resolver does not die, is a predicate on behaviors. § In this Section §
5, we will assume that S is not the set of everything,and, in particular, we will assume that there is not any function in S . We will focus onthe survival problem, and then assume that S is the set of behaviors, or finite stateautomata, but you can think that S = N , or any other countable set, see § S , to more complex resolvers that have to implement functions in order tolook for resolutions to deal with metaproblems. § A problem type, for example P Ψ , is a subset of the set of problems,that is, P Ψ ⊆ P . We will note S Ψ the set of the solutions to the type of problems P Ψ .That is, ∀ π Ψ ∈ P Ψ , Σ π Ψ ⊆ S Ψ ⊆ S . Comment
The survival problem is not a single problem, but a type of problems, P Ω ;each living being faces a different survival problem. But, in this case as in manyothers, what it is certain is that the solutions to any of these problems is of a specifickind. For example, while eating can be a solution, imagining how to eat is not asolution, even though it can help us to get something to eat, because it can be ametasolution. Then S Ω is the set of behaviors. § Metaproblems Π π are a type of problem, Π P = P Π , and its solutionsare resolutions, Π S = S Π = R , see § § If the set of the solutions to some type of problem is finite, 0 < | S Γ | < ℵ ,then each and every problem of that type is expressible and resolvable. Proof
Because those problems are in F , so Lemma § ⋄ Comment
If 0 < | S Γ | = N < ℵ , then | P Γ | = 2 N < ℵ and | R Γ | = (2 N ) N < ℵ . Inthe finite case, | S Γ | < | P Γ | < | R Γ | < ℵ . § A constant function K s : S → S is: ∀ s ∈ S , ∀ x ∈ S , K s ( x ) = s . Comment
Every constant function K s is effectively calculable, see § λ -definable; in λ -calculus, K = ( λsx.s ). This is because our λ -calculus includes the K combinator, and so we refer to the λK -calculus simply as λ -calculus. Special cases
Tautology: K ⊤ = P τ . Contradiction: K ⊥ = P ¯ τ . See § § The constant isomorphism is the natural isomorphism between S and the set of constant functions K that relates each s ∈ S with K s ∈ ( S → S ). Thatis, S ⇔ K : s ↔ K s . Comment
We can extend any operation on S to K . For example, for any binaryoperation ∗ on S , we define ∀ x, [ K a ∗ K b ]( x ) = K a ( x ) ∗ K b ( x ) = a ∗ b = K a ∗ b ( x ). Comment
Semantics is included in syntax, that is, S ∼ = K ⊂ ( S → S ). § A semantic function f : S → S is a syntactic element, f ∈ ( S → S ),but it is not a syntactic function f ∈ (( S → S ) → ( S → S )), because the semanticfunction f takes semantic elements and returns semantic elements, while, using theconstant isomorphism, the syntactic function f is not restricted. In particular, asemantic function cannot take a function, and a semantic function cannot return afunction. ww.ramoncasares.com PT 31
Comment
In semantics, literal identity i is the identity, see § f ( x ) = y , which means that f ( x ) and y aretwo syntactic objects that refer to the same semantic object. Then, there are twoidentities in syntax: literal identity i , or quotation, which is the semantic functionthat just returns what it takes, and functional identity u , or evaluation, which is thesyntactic function that follows the references and returns the final one, see § § The range of a resolver ℜ , noted Ξ ℜ , is the set of the problemsfor which ℜ provides a non-empty subset of solutions, and only of solutions, that is,Ξ ℜ = { π | ℜ ( π ) ⊆ Σ π ∧ ℜ ( π ) = ∅ } . Comment
The range of a resolver is the set of the problems that the resolver solves. § The power of a resolver ℜ , noted Φ ℜ , is the set of the problemsthat the resolver ℜ resolves, that is, Φ ℜ = { π | ℜ ( π ) = Σ π } . Comment
In practice, if | Σ π | >
1, it is not sensible to generate all the solutions, Σ π ,when just one solution solves the problem. In these cases the range of the resolver ismore important than its power. § ∀ℜ , S ∩ Φ ℜ ⊆ Ξ ℜ . Proof
Because ∀ π ∈ Φ ℜ ∩ S , we have that π ∈ Φ ℜ so ℜ ( π ) = Σ π , see § ℜ ( π ) ⊆ Σ π , and also that π ∈ S , so | Σ π | > ⇔ Σ π = ∅ , see § § ℜ ( π ) = Σ π = ∅ , and therefore π ∈ Ξ ℜ , see § ⋄ Comment
For solvable problems, Φ
ℜ ⊆ Ξ ℜ , so they are easier to solve than to resolve.But unsolvable problems, some of them resolved, are impossible to solve! § We will say that the resources of a resolver are in a set if thecapability implemented in the resolver belongs to that set.
Comment
Now we will construct a series of resolvers ℜ n , from the minimal one thatonly implements one solution, and then growing naturally step by step. Each resolverwill implement just one element out of its resources Notation
We will use {· ℜ n ·} to refer to the set of all the resolvers of step n . § § A mechanism ℜ is any resolver that implements one member of S . We will note ℜ [ s ], where s ∈ S , the mechanism that implements s , that is, ℜ [ s ] = s ∈ S . Then the mechanisms resources are in S , and {· ℜ ·} = S . Comment
Mechanism ℜ [ s ] returns s unconditionally. Comment
A mechanism ℜ implements a semantic unconditional computation. § As resolutions return sets of elements in S , to normalize the situationof mechanisms ℜ we will use the singleton isomorphism, see § ℜ [ { s } ] to mean the singleton { s } , that is, ℜ [ { s } ] = {ℜ [ s ] } = { s } ∈ S . ww.ramoncasares.com PT 32 § ∀ s ∈ S , Ξ ℜ [ { s } ] = { π | P π ( s ) } . Proof
Just applying the definition of range, see § ℜ [ { s } ] = { π | ℜ [ { s } ] ⊆ Σ π ∧ ℜ [ { s } ] = ∅ } = { π | { s } ⊆ Σ π ∧ { s } 6 = ∅ } = { π | s ∈ Σ π ∧ ⊤ } = { π | s ∈ Σ π } = { π | P π ( s ) } . ⋄ Comment
The range of the mechanism ℜ [ s ] is the set of problems for which s is asolution. § ∀ s ∈ S , Φ ℜ [ { s } ] = { δ s } . Proof
Just applying the definition of power, see § ℜ [ { s } ] = { π | ℜ [ { s } ] = Σ π } = { π | { s } = Σ π } = { δ s } , the last equationbecause Σ δ s = { s } , see § ⋄ Comment
Mechanism ℜ [ s ] only resolves problem δ s . § Any singleton routine resolution R π = { s } can be implemented by themechanism ℜ [ R π ]. Proof If R π = { s } , then R π = { s } = {ℜ [ s ] } = ℜ [ { s } ] = ℜ [ R π ]. ⋄ Comment
In theory, we can equal any finite routine resolution to a union of a finitenumber of mechanisms, R π = Σ π = S s ∈ Σ π { s } = S s ∈ Σ π {ℜ [ s ] } . § In practice, it only makes sense to implement one solution, as ℜ [ s ]does. Without conditional calculations, the mechanism could not control when toapply one result or any of the others, so it would gain nothing implementing morethan one. Comment
The mechanism is a body capable of one behavior.
Example
A mechanism can only survive in a specific and very stable environment, asit is the case of some extremophile archaea. § § An adapter ℜ is any resolver that implements one condition onthe members of S . We will note ℜ [ P S ] the adapter that implements P S , where P S ∈ ( S → B ), that is, ℜ [ P S ] = P S ∈ ( S → B ). Then the adapters resources are in S → B , and {· ℜ ·} = ( S → B ). Comment
An adapter ℜ implements a semantic conditional computation. § Each adapter ℜ [ P S ] implements one set of elements of S . Proof
Because every predicate P S defines a set S = { s ∈ S | P S ( s ) } ∈ S . Thecondition P S is the characteristic function of S , ∀ s ∈ S , P S ( s ) = [ s ∈ S ]. ⋄ Comment
We will write ℜ [ P S ] = ℜ [ S ] = S ∈ S . Only effectively calculable con-ditions are implementable, and then adapters can only implement expressible, orrecursive, sets, E . Then, ℜ [ P S ] = ℜ [ S ] = S ∈ E . § Every mechanism ℜ is an adapter ℜ , that is, {· ℜ ·} ⊂ {· ℜ ·} . Proof
For each mechanism ℜ [ s ], which implements s ∈ S , there is an adapter ℜ [ P δ s ],see § { s } ∈ ( S → B ). But not every set is asingleton. Summarizing, {· ℜ ·} = S ⊂ ( S → B ) = {· ℜ ·} . ⋄ Comment
In Cantor’s paradise, but out of Turing universes, by the singleton ( § § {· ℜ ·} = S ∼ = S ⊂ S ∼ = ( S → B ) = {· ℜ ·} . ww.ramoncasares.com PT 33 § ℜ [ S ] = S s ∈ S {ℜ [ s ] } . Proof
Because ℜ [ s ] = s , so S s ∈ S {ℜ [ s ] } = S s ∈ S { s } = S = ℜ [ S ]. ⋄ Comment
The results are the same, but not the implementation, because while theadapter ℜ [ S ] implements a condition, the union of mechanisms S s ∈ S {ℜ [ s ] } worksunconditionally. Thus, the output of the union of mechanisms is independent ofany problem, and then the union cannot implement ℜ [ P S ∧ P π ] = ℜ [ S ∩ Σ π ], forexample, so it cannot implement any trial, see Theorem § § ∀ S ∈ S , ∀ π ∈ P , Ξ ℜ [ S ∩ Σ π ] = { π | S ∩ Σ π = ∅ } . Proof
Because Ξ ℜ [ S ∩ Σ π ] = { π | ( ℜ [ S ∩ Σ π ] ⊆ Σ π ) ∧ ( ℜ [ S ∩ Σ π ] = ∅ ) } = { π | ( S ∩ Σ π ⊆ Σ π ) ∧ ( S ∩ Σ π = ∅ ) } = { π | ⊤ ∧ ( S ∩ Σ π = ∅ ) } = { π | S ∩ Σ π = ∅ } . ⋄ Comment
The range of the adapter ℜ [ S ∩ Σ π ] is the set of problems that have anysolution in S . The adapter ℜ [ S ∩ Σ π ] solves any problem such that any of its solutionsare in S . Corollary If S ⊂ S ′ , then Ξ ℜ [ S ∩ Σ π ] ⊂ Ξ ℜ [ S ′ ∩ Σ π ]. Proof
In that case, if a solution to a problem is in S , then it is also in S ′ . But thereare also solutions in S ′ that are not in S . ⋄§ If s ∈ S , then ∀ π ∈ P , Ξ ℜ [ { s } ] ⊆ Ξ ℜ [ S ∩ Σ π ]. Proof
By Lemma § ∀ π ∈ Ξ ℜ [ { s } ], P π ( s ), that is, s ∈ Σ π , so, if s ∈ S , then S ∩ Σ π = ∅ , and therefore π ∈ Ξ ℜ [ S ∩ Σ π ], by Lemma § ⋄ Definition
We will call s ∈ S the adapter condition. If the adapter condition holds,then the adapter ℜ [ S ∩ Σ π ] solves any problem that the mechanism ℜ [ s ] solves. Corollary If { s } ⊂ S , then Ξ ℜ [ { s } ] ⊂ Ξ ℜ [ S ∩ Σ π ]. Proof
Because, if t ∈ S and t = s , then δ t ∈ Ξ ℜ [ S ∩ Σ π ] but δ t Ξ ℜ [ { s } ]. ⋄ Proposition If { s } ⊂ S , then Ξ ℜ [ { s } ] Ξ ℜ [ S ].Because δ s ∈ Ξ ℜ [ { s } ], but δ s Ξ ℜ [ S ]. § ∀ S ∈ S , ∀ π ∈ P , Φ ℜ [ S ∩ Σ π ] = 2 S . Proof
Because Φ ℜ [ S ∩ Σ π ] = { π | ℜ [ S ∩ Σ π ] = Σ π } = { π | S ∩ Σ π = Σ π } = { π | Σ π ⊆ S } = 2 S , where the last equality uses the set isomorphism, see § ⋄ Comment
The power of the adapter ℜ [ S ∩ Σ π ] is the powerset of S . The adapter ℜ [ S ∩ Σ π ] resolves any problem such that all of its solutions are in S . Corollary If S ⊂ S ′ , then Φ ℜ [ S ∩ Σ π ] ⊂ Φ ℜ [ S ′ ∩ Σ π ]. Proof
Just because, if S ⊂ S ′ , then 2 S ⊂ S ′ . ⋄§ If s ∈ S , then ∀ π ∈ P , Φ ℜ [ { s } ] ⊂ Φ ℜ [ S ∩ Σ π ]. Proof
Using the set isomorphism, see § δ s = Σ δ s = { s } , and then, if s ∈ S ,Φ ℜ [ { s } ] = { δ s } = ⊂ S = Φ ℜ [ S ∩ Σ π ], by Lemmas § § ⋄ Comment
If the adapter condition holds, s ∈ S , then the adapter ℜ [ S ∩ Σ π ] resolvesany problem that the mechanism ℜ [ s ] resolves, and more. Proposition If { s } ⊂ S , then Φ ℜ [ { s } ] Φ ℜ [ S ], because δ s Φ ℜ [ S ] = { S } . § Any effectively calculable trial resolution T π ( S ) can be implemented bythe adapter ℜ [ S ∩ Σ π ]. Proof T π ( S ) = { s ∈ S | s ∈ Σ π } = { s | s ∈ S ∧ s ∈ Σ π } = { s | P S ( s ) ∧ P π ( s ) } . Then T π ( S ) . = ℜ [ P S ∧ P π ] = ℜ [ S ∩ Σ π ]. The equality is dotted because, if the trial is notan effectively calculable function, then it cannot be implemented. ⋄ ww.ramoncasares.com PT 34 § In practice, an adapter ℜ [ P S ∧ P π ] = ℜ [ S ∩ Σ π ] has a body capableof several behaviors that provides the set S of behaviors. If the current behavior werenot satisfying the adapter condition P π , which is interpreted as an error, then theadapter would change its behavior trying another one in S . Comment
The adapter is a body capable of several behaviors, and a governor thatselects the current behavior.
Example
A deciduous tree, which switches its behavior with seasons, is an adapter. § § A perceiver ℜ is any resolver that implements one transformationof the elements in S into the elements in S . We will note ℜ [ f ] the perceiver thatimplements f , where f ∈ ( S → S ), that is, ℜ [ f ] = f ∈ ( S → S ). Then the perceiverresources are in S → S , and {· ℜ ·} = ( S → S ). Comment
From a semantic point of view, a perceiver ℜ implements a semantic func-tional computation. From a syntactic point of view, a perceiver ℜ implements asyntactic unconditional computation. § Perceivers are to syntax as mechanisms are to semantics.
Comment
When solutions are functions S → S , then a perceiver does what a mech-anism does, which is to return a solution unconditionally. That is, perceivers onmetaproblems are as mechanisms on problems. But, perceivers can go further. Comment
The perceiver ℜ [ f ] implements function f from S to S , that is, f : S → S .Then, ∀ s ∈ S , ℜ [ f ]( s ) = f ( s ) ∈ S . § By the rewriting rules in § ℜ [ f ]( S ) = { ℜ [ f ]( s ) | s ∈ S } ∈ S .Then ℜ [ f ]( S ) returns a set of solutions, as any well-behaved resolution should do. Comment
The perceiver ℜ [ f ]( S ) implements f , meaning that f is hardwired in theperceiver, while S is just data. We will call what is implemented hardware, andwhat is data software. We write the hardware between brackets, and the software be-tween parentheses. We will assume that coding software costs less than implementinghardware, or, in fewer words, that software is cheaper than hardware § Every adapter ℜ is a perceiver ℜ , that is, {· ℜ ·} ⊂ {· ℜ ·} . Proof
Because B ⊂ S , and then ( S → B ) ⊂ ( S → S ). So {· ℜ ·} = ( S → B ) ⊂ ( S → S ) = {· ℜ ·} . ⋄ Comment
Each adapter implements one condition P S ∈ ( S → B ). And any condition P S ∈ ( S → B ) is also a function P S ∈ ( S → S ), because B ⊂ S . Therefore, for eachadapter ℜ [ P S ], which implements condition P S , there is a perceiver ℜ [ P S ] thatimplements the function P S , and then we write ℜ [ P S ] = P S = ℜ [ P S ]. Comment
Again, ℜ [ P S ] = ℜ [ P S ] explains that the results are the same, but not theimplementation. § ∀ S ∈ S , ℜ [ S ] = ℜ [ i ]( S ). Proof
Function i : S → S is the semantic identity, i = i see § ∀ s ∈ S , i ( s ) = s ,and ℜ [ i ]( S ) = { ℜ [ i ]( s ) | s ∈ S } = { i ( s ) | s ∈ S } = { s | s ∈ S } = S = ℜ [ S ]. ⋄ Comment
The same perceiver hardware ℜ [ i ], just by changing its software, can em-ulate different adapters: ℜ [ i ]( S ) = ℜ [ S ], and ℜ [ i ]( S ′ ) = ℜ [ S ′ ]. Then the per-ceiver ℜ [ i ]( S ) is more flexible than the adapter ℜ [ S ], because S is hardwired in theadapter, while it is easily replaceable data for the perceiver. ww.ramoncasares.com PT 35 § ∀ S ∈ S , Ξ ℜ [ S ] = Ξ ℜ [ i ]( S ) and Φ ℜ [ S ] = Φ ℜ [ i ]( S ). Proof
Because, by Lemma § ℜ [ S ] = ℜ [ i ]( S ). ⋄ Definition
The perceiver condition is satisfied if it implements the semantic identity i . Comment
If the perceiver condition holds, then the perceiver ℜ [ i ]( S ) solves any prob-lem solved by the adapter ℜ [ S ], and the perceiver ℜ [ i ]( S ) resolves any problemresolved by the adapter ℜ [ S ]. Remark
Semantic identity i is the ideal for perception. § Ξ ℜ [ i ]( S ∩ Σ π ) = Ξ ℜ [ S ∩ Σ π ] and Φ ℜ [ i ]( S ∩ Σ π ) = Φ ℜ [ S ∩ Σ π ]. Proof
By Lemma § ⋄ Comment
The same perceiver hardware ℜ [ i ] can be tuned to a different trial just bychanging its software, from ℜ [ i ]( S ∩ Σ π ) to ℜ [ i ]( S ′ ∩ Σ ρ ), for example. Proposition If S ⊂ S ′ , then Ξ ℜ [ i ]( S ∩ Σ π ) ⊂ Ξ ℜ [ i ]( S ′ ∩ Σ π ), and Φ ℜ [ i ]( S ∩ Σ π ) ⊂ Φ ℜ [ i ]( S ′ ∩ Σ π ), by Lemma § § § § A function on sets F : 2 S → S is elementable if it exists aneffectively calculable function f : S → S such that ∀ S, F ( S ) = { f ( s ) | s ∈ S } . Comment
We write F ( S ) = f ( S ), by the rules in § f is a semantic function, f : S → S , that f is effectively calculable, and that F ( S ) = f ( S ). Proposition
Set identity I : 2 S → S | ∀ S ∈ S , I ( S ) = S , is elementable by semanticidentity i , because ∀ S ∈ S , i ( S ) = { i ( s ) | s ∈ S } = { s | s ∈ S } = S = I ( S ). § Any analogy resolution A ◦ T Aπ ( S ) ◦ T A can be implemented by thetri-perceiver ℜ [ T a ]( ℜ [ i ]( S ∩ ℜ [ a ](Σ π ))), if A is elementable by a , and T A by T a . Proof
An analogy resolution is A ◦ T Aπ ( S ) ◦ T A . Both A and T A are functions fromsets to sets, T A : 2 S → S and A : ( P → P ) = (2 S → S ), so if both A and T A are elementable, then a perceiver can implement them. We have a and T a , whichare both semantic functions such that A ( S ) = a ( S ), and T A ( S ) = T a ( S ). Then P aπ = a ( P π ) = ℜ [ a ]( P π ) = ℜ [ a ](Σ π ) implements the first third, T aπ ( S ) = ℜ [ P S ∧ P aπ ] = ℜ [ P S ∧ ℜ [ a ]( P π )] = ℜ [ S ∩ ℜ [ a ](Σ π )] implements the second third, see § § ℜ [ T a ]( ℜ [ i ]( S ∩ ℜ [ a ](Σ π ))) implements the whole analogyresolution a ◦ T aπ ( S ) ◦ T a . ⋄ Corollary
Identity analogy I ◦ T Iπ ( S ) ◦ T I can be implemented by the bi-perceiver ℜ [ i ]( S ∩ ℜ [ i ](Σ π )), which uses the identity perceiver ℜ [ i ] twice. Proof ℜ [ T i ]( ℜ [ i ]( S ∩ℜ [ i ](Σ π ))) = ℜ [ i ]( ℜ [ i ]( S ∩ℜ [ i ](Σ π ))) = ℜ [ i ]( S ∩ℜ [ i ](Σ π )),because set identity I is elementable by semantic identity i , and T I = I , so it is alsoelementable by T i = i , and i ◦ i = i . ⋄§ While an adapter uses a trial and error resolution, and this meansthat error is part of the usual procedure, a perceiver executes the trial and error insideitself. If the analogy provides a good model, then the internal trial is as good as theexternal one, with the advantage that the errors are only simulated errors. More tothe point, if the problem the resolver faces is the survival problem, then the adaptererrors are literally death errors, or at least pain, while the perceiver errors are justmental previsions of what not to do. See that, if the perceiver implements the identityanalogy, as ℜ [ i ] does, then the model is good, because the internal problem is equalto the external one. And the perceiver ℜ [ i ] is more flexible than the adapter. ww.ramoncasares.com PT 36
Comment
The perceiver is a body capable of several behaviors, a governor that selectsthe current behavior, and a simulator that internalizes behaviors.
Example
The perceiver governor determines what to do based upon an internal in-terpretation. According to Lettvin et al. (1959), a frog is a perceiver that uses aninternal routine. Frog’s i is such that any dark point that moves rapidly in its fieldof vision is a fly which it will try to eat. § § A learner ℜ is any resolver that implements one condition on themembers of S → S . We will note ℜ [ P F ] the learner that implements P F , where P F ∈ (( S → S ) → B ), that is, ℜ [ P F ] = P F ∈ (( S → S ) → B ). Then the learnersresources are in ( S → S ) → B , and {· ℜ ·} = (( S → S ) → B ). Comment
A learner ℜ implements a syntactic conditional computation. § Learners are to syntax as adapters are to semantics.
Comment
When solutions are functions S → S , then a learner does what an adapterdoes, which is to return a predicate on solutions. That is, learners on metaproblemsare as adapters on problems. But, learners can go further. § Each learner ℜ [ P F ] implements one set of members of ( S → S ). Proof
Because every predicate P F : ( S → S ) → B defines a set F = { f ∈ ( S → S ) | P F ( f ) } ∈ S → S . The condition P F is the characteristic functionof F , ∀ f ∈ ( S → S ) , P F ( f ) = [ f ∈ F ]. ⋄ Comment
We will write ℜ [ P F ] = ℜ [ F ] = F ∈ S → S . Comment
The learner ℜ [ F ] implements F ∈ S → S . So ∀ s ∈ S , ℜ [ F ]( s ) = F ( s ) ∈ S ,because F ( s ) = { f ( s ) | f ∈ F } , by the rewriting rules in § ℜ [ F ]( s )returns a set of solutions, as any well-behaved resolution should do. Also, by thesame rules, ℜ [ F ]( S ) = F ( S ) = { f ( s ) | s ∈ S × f ∈ F } ∈ S . § Every perceiver ℜ is a learner ℜ , that is, {· ℜ ·} ⊂ {· ℜ ·} . Proof
For each perceiver ℜ [ f ], which implements f ∈ ( S → S ), there is a learner ℜ [ P δ f ], see § { f } ∈ (( S → S ) → B ). But notevery set is a singleton. Then, {· ℜ ·} = ( S → S ) ⊂ (( S → S ) → B ) = {· ℜ ·} . ⋄§ ℜ [ F ] = S f ∈ F {ℜ [ f ] } . Proof
Because ℜ [ f ] = f , so S f ∈ F {ℜ [ f ] } = S f ∈ F { f } = F = ℜ [ F ]. ⋄ Comment
Again, the results are the same, but not the implementation. The union ofperceivers S f ∈ F {ℜ [ f ] } cannot select a function to use, so it cannot implement anymeta-trial, as ℜ [ R ∩ Σ Π π ] does, see § Comment
These are not sets of solutions, but sets of semantic functions. § ∀ f ∈ F , ∀ S ∈ S , ℜ [ f ]( S ) ⊆ ℜ [ F ]( S ). Proof If f ∈ F , then f ( S ) = { f ( s ) | s ∈ S } ⊆ { f ′ ( s ) | s ∈ S × f ′ ∈ F } = F ( S ). ⋄§ We will rewrite ℜ [ ℜ [ F ]( S ) ∩ Σ π ] as ℜ [ F ]( S ∩ ∩ Σ π ). Comment
We can write ℜ [ ℜ [ F ]( S ) ∩ Σ π ] = ℜ [ F ]( S ∩ ∩ Σ π ), because any learner canimplement semantic conditions, that is, because {· ℜ ·} ⊂ {· ℜ ·} . ww.ramoncasares.com PT 37 § If f ∈ F , then ∀ S ∈ S , ∀ π ∈ P , Ξ ℜ [ f ]( S ) ⊆ Ξ ℜ [ F ]( S ∩ ∩ Σ π ). Proof
Firstly see that, if f ∈ F , then f ( S ) ⊆ F ( S ), by § f ( S ) ∩ Σ π ⊆ F ( S ) ∩ Σ π . Secondly see that ∀ π ∈ Ξ ℜ [ f ]( S ), f ( S ) ∩ Σ π = f ( S ) = ∅ . This isbecause Ξ ℜ [ f ]( S ) = { π | f ( S ) ⊆ Σ π ∧ f ( S ) = ∅ } . Now, taking both together, ∀ π ∈ Ξ ℜ [ f ]( S ), ∅ 6 = f ( S ) = f ( S ) ∩ Σ π ⊆ F ( S ) ∩ Σ π , so for these π , F ( S ) ∩ Σ π = ∅ .Then these π ∈ { π | F ( S ) ∩ Σ π ⊆ Σ π ∧ F ( S ) ∩ Σ π = ∅ } = Ξ ℜ [ ℜ [ F ]( S ) ∩ Σ π ],because F ( S ) ∩ Σ π ⊆ Σ π is always true. ⋄ Definition
We will call f ∈ F the learner condition. If the learner condition holds,then the learner ℜ [ F ]( S ∩ ∩ Σ π ) solves any problem that the perceiver ℜ [ f ]( S ) solves. § If f ∈ F , then ∀ S ∈ S , ∀ π ∈ P , Φ ℜ [ f ]( S ) ⊆ Φ ℜ [ F ]( S ∩ ∩ Σ π ). Proof If f ∈ F , then f ( S ) ⊆ F ( S ), see § f ( S ) ∩ Σ π ⊆ F ( S ) ∩ Σ π . Now ∀ π ∈ Φ ℜ [ f ]( S ), f ( S ) = Σ π , and then for these π , Σ π = f ( S ) ∩ Σ π ⊆ F ( S ) ∩ Σ π ⊆ Σ π .Therefore, for these π , F ( S ) ∩ Σ π = Σ π , and Φ ℜ [ f ]( S ) ⊆ { π | F ( S ) ∩ Σ π = Σ π } =Φ ℜ [ ℜ [ F ]( S ) ∩ Σ π ]. ⋄ Comment
If the learner condition holds, f ∈ F , then the learner ℜ [ F ]( S ∩ ∩ Σ π ) resolvesany problem that the perceiver ℜ [ f ]( S ) resolves. § In particular, if i ∈ R , then Ξ ℜ [ i ]( S ∩ Σ π ) ⊆ Ξ ℜ [ R ]( S ∩ ∩ Σ π )and Φ ℜ [ i ]( S ∩ Σ π ) ⊆ Φ ℜ [ R ]( S ∩ ∩ Σ π ). Proof
By Lemmas § § ℜ [ R ]( S ∩ Σ π ∩ ∩ Σ π ) ⊆ Ξ ℜ [ R ]( S ∩ ∩ Σ π ),and Φ ℜ [ R ]( S ∩ Σ π ∩ ∩ Σ π ) ⊆ Φ ℜ [ R ]( S ∩ ∩ Σ π ), because ℜ [ R ]( S ∩ Σ π ) ⊆ ℜ [ R ]( S ), socorollaries to Lemmas § § ⋄§ Any meta-trial resolution T Π π ( R ) can be implemented by the learner ℜ [ R ∩ Σ Π π ], if P R and P Π π are elementable. Comment
The diagram for the meta-trial, or trial of the metaproblem, see § π Π −−→ Π π T Π π ( R ) −−−−−→ Σ Π π T Π −−→ Σ π . Comment
A learner solves metaproblems by trial, as an adapter solves problems bytrial. The following correlations stand: ℜ ↔ ℜ , Π π ↔ π , R ↔ S , and then ℜ [ R ∩ Σ Π π ] ↔ ℜ [ S ∩ Σ π ]. Therefore, ℜ [ R ∩ Σ Π π ] compares to ℜ [ i ], where i ∈ R ,as ℜ [ S ∩ Σ π ] compares to ℜ [ s ], where s ∈ S . Proof T Π π ( R ) = { r ∈ R | r ∈ Σ Π π } = { r | r ∈ R ∧ r ∈ Σ Π π } = { r | P R ( r ) ∧ P Π π ( r ) } .In the meta-trial T Π π ( R ), R is a set of resolutions, where R = ( P → S ) = (2 S → S ),that is, P R : (2 S → S ) → B , and the condition of the metaproblem Π π is also P Π π : R → B = ( P → S ) → B = (2 S → S ) → B . So if both P R and P Π π are elementableby ℘ R and ℘ Π π , then both of them, and its conjunction, are implementable. Then T Π π ( R ) = { r | P R ( r ) ∧ P Π π ( r ) } . = ℜ [ ℘ R ∧ ℘ Π π ] = ℜ [ R ∩ Σ Π π ]. ⋄§ Perceiver success depends crucially on the analogy, that is, on howmuch the analogy resembles the identity i . And a learner can adapt the analogy tothe problem it is facing, because the learner ℜ [ R ] implements a set of functions R from which it can select another analogy when the current one fails. Adapting theanalogy is also known as modeling. So a learner can apply different analogies, but alearner can also apply a routine if it knows a solution, because the routine is moreefficient, or a trial, when the model is not good enough or too pessimistic. ww.ramoncasares.com PT 38
Comment
The learner is a body capable of several behaviors, a governor that selectsthe current behavior, a simulator that internalizes behaviors, and a modeler thatadjusts the model used by the simulator.
Example
Where there is modeling and simulation there is learning, because enhancingthe model prevents repeating errors. A dog is a learner. § § A subject ℜ is any resolver that implements one transformationof the elements in S → S into the elements in S → S . We will note ℜ [ f ] the subjectthat implements f , where f ∈ (( S → S ) → ( S → S )), that is, ℜ [ f ] = f ∈ (( S → S ) → ( S → S )). Then the subject resources are in ( S → S ) → ( S → S ), and {· ℜ ·} = (( S → S ) → ( S → S )). Comment
A subject ℜ implements a syntactic functional computation. § Subjects are to syntax as perceivers are to semantics.
Comment
When solutions are functions S → S , then a subject does what a perceiverdoes, which is to return a function on solutions to solutions. That is, subjects onmetaproblems are as perceivers on problems. But, subjects can go further. Comment
The subject ℜ [ f ] implements function f from S → S to S → S , that is, f : ( S → S ) → ( S → S ). Then, ∀ f ∈ ( S → S ) , ℜ [ f ]( f ) = f ( f ) ∈ ( S → S ). § As resolutions return sets of elements in S , to normalize the situationof subjects, for which ℜ [ f ]( f )( s ) ∈ S , we will use the rewriting rules in § ℜ [ f ]( F )( S ) = { ℜ [ f ]( f )( s ) | s ∈ S × f ∈ F } ∈ S . Comment
Subject ℜ [ f ]( F )( S ) has two software levels: semantics ( S ) and syntax ( F ). § Every learner ℜ is a subject ℜ , that is, {· ℜ ·} ⊂ {· ℜ ·} . Proof
First we define set B = { K ⊤ , K ⊥ } , both functions S → S , see § B and B , mapping ⊤ to the functionthat always returns ⊤ , which is K ⊤ = P τ , and ⊥ to the function that always returns ⊥ , which is K ⊥ = P ¯ τ , see § B ⇔ B : ⊤ ↔ K ⊤ , ⊥ ↔ K ⊥ , and B ∼ = B ⊂ ( S → S ). Then, (( S → S ) → B ) ⊂ (( S → S ) → ( S → S )), and therefore {· ℜ ·} = (( S → S ) → B ) ⊂ (( S → S ) → ( S → S )) = {· ℜ ·} . ⋄ Comment
Each learner implements one condition P F ∈ (( S → S ) → B ). And anycondition on functions P F ∈ (( S → S ) → B ) is also a function on functions tofunctions P F ∈ (( S → S ) → ( S → S )), because B ⊂ ( S → S ). Therefore, foreach learner ℜ [ P F ], which implements condition P F , there is a subject ℜ [ P F ] thatimplements the function P F , and then we write ℜ [ P F ] = P F = ℜ [ P F ]. Comment
Again, ℜ [ P F ] = ℜ [ P F ] explains that the results are the same, but not theimplementation. § ∀ F ∈ S → S , ∀ S ∈ S , ℜ [ F ]( S ) = ℜ [ u ]( F )( S ). Comment
Function u is the identity for programs, see § § u is syntactic because it is not restricted to semanticobjects, see § u is equivalent to λ -calculus I = ( λx.x ), see § ∀ f ∈ S → S , u ( f ) = f and ∀ s ∈ S , u ( f )( s ) = f ( s ). Proof
By the rewriting rules in § ℜ [ u ]( F )( S ) = { ℜ [ u ]( f )( s ) | s ∈ S × f ∈ F } = { u ( f )( s ) | s ∈ S × f ∈ F } = { f ( s ) | s ∈ S × f ∈ F } = F ( S ) = ℜ [ F ]( S ). ⋄ Corollary ∀ F ∈ S → S , ℜ [ F ] = ℜ [ u ]( F ). ww.ramoncasares.com PT 39 § ∀ F ∈ S → S , ∀ S ∈ S , Ξ ℜ [ F ]( S ) = Ξ ℜ [ u ]( F )( S ) andΦ ℜ [ F ]( S ) = Φ ℜ [ u ]( F )( S ). Proof
Because, by Lemma § ℜ [ F ]( S ) = ℜ [ u ]( F )( S ). ⋄ Definition
The subject condition is satisfied if it implements the functional identity u . Comment
If the subject condition holds, then the subject ℜ [ u ]( F )( S ) solves any prob-lem solved by the learner ℜ [ F ]( S ), and also the subject ℜ [ u ]( F )( S ) resolves anyproblem resolved by the learner ℜ [ F ]( S ). Remark
Functional identity u is the ideal for reason. § In particular, Ξ ℜ [ u ]( R )( S ∩ ∩ Σ π ) = Ξ ℜ [ R ]( S ∩ ∩ Σ π ) andΦ ℜ [ u ]( R )( S ∩ ∩ Σ π ) = Φ ℜ [ R ]( S ∩ ∩ Σ π ). Proof
By Lemma § ⋄ Comment
Subject ℜ [ u ]( R )( S ∩ ∩ Σ π ) is more flexible than learner ℜ [ R ]( S ∩ ∩ Σ π ), because R is software for the subject while it is hardware in the learner, and software is cheaperthan hardware, see § § Subject ℜ [ u ] is a full resolution machine. Proof
By Theorem § § ℜ [ u ] = u = c ( U ), so using the programisomorphism, see § ℜ [ u ] = U , which is a Turing complete device, and thereforeis a full resolution machine, by Theorem § ⋄§ Any effectively calculable resolution ℜ can be implemented by the sub-ject ℜ [ u ]. Proof
By Theorem § ⋄ Corollary
Any effectively calculable meta-analogy resolution
A ◦ T A Π π ( R ) ◦ T A canbe implemented by the subject ℜ [ u ], including metaresolving, see § § The subject, by internalizing metaproblems, prevents meta-errors,that is, the subject can test internally a resolution before executing it. The subjectis also more flexible than the learner, because subject modeling is done in software,instead of in hardware. And subject ℜ [ u ] can reason about any model. This meansthat subject ℜ [ u ] is a resolver that can calculate solutions, but also problems andresolutions without limits; it can represent the problem it is facing to itself, and itcan represent itself to itself. In this sense, the subject ℜ [ u ] is conscious. Comment
The subject is a body capable of several behaviors, a governor that selectsthe current behavior, a simulator that internalizes behaviors, a modeler that adjuststhe model used by the simulator, and a reason that internalizes resolutions.
Example
It seems that only our species,
Homo sapiens , is Turing complete. We dealwith the evolution to Turing completeness and its relation to language in Casares(2016b). § § There is a hierarchy of resolvers: {· ℜ ·} ⊂ {· ℜ ·} ⊂ {· ℜ ·} ⊂ {· ℜ ·} ⊂ {· ℜ ·} . Proof
Because S ⊂ ( S → B ) ⊂ ( S → S ) ⊂ (( S → S ) → B ) ⊂ (( S → S ) → ( S → S )),by Lemmas § § § § ℜ [ u ] = U , by Theorems § § ⋄ ww.ramoncasares.com PT 40 § This table groups concepts closely related from problem theory, astrial, computing theory, as adapter ℜ , and set theory, as S ∈ S → B . Semantics Syntax
Routine Meta-routineone Mechanism ℜ Perceiver ℜ element s ∈ S f ∈ ( S → S )Trial Meta-trialsome Adapter ℜ Learner ℜ set S ∈ S → B F ∈ ( S → S ) → B Analogy Meta-analogyany Perceiver ℜ Subject ℜ function f ∈ S → S f ∈ ( S → S ) → ( S → S ) S S → S Elements Functions
Comment
A perceiver is a syntactic mechanism. A learner is a syntactic adapter. Asubject is a syntactic perceiver. A subject is a syntactic mechanism. § The problem theory is complete.
Proof
Aside from definitions, the problem theory posits that there are three ways toresolve a problem: routine, trial, and analogy; see § § § § § § § § § S → S → S → S , see § § S → S ) → ( S → S ). Andthere is a subject that is a Turing complete device, ℜ [ u ], see § § § ⋄ Comment
It also means that no more resolutions are needed, although we could dowithout routine, for example, by using Theorem § § Comment
This theorem is true if the Turing’s thesis is true, see § § Provided that a bigger range means more survival opportunities, thatsoftware is cheaper than hardware, that the adapter, the perceiver, the learner, andthe subject conditions are satisfied in some environments, and that in each step theincreasing of complexity was overcome by its fitness, then an evolution of resolvers—mechanism to adapter to perceiver to learner to subject— should follow. ww.ramoncasares.com
PT 41
Comment
Although depending on conditions, see Lemmas § § § § ℜ [ u ]. Comment
In detail, the strictest evolution of resolvers is: Ξ ℜ [ s ] { s }⊂ S ⊂ Ξ ℜ [ S ∩ Σ π ] S ⊂ S ′ ⊂ Ξ ℜ [ i ]( S ′ ∩ Σ π ) { i }⊂ R ⊆ Ξ ℜ [ R ]( S ′ ∩ ∩ Σ π ) R ⊂ R ′ ⊆ Ξ ℜ [ u ]( R ′ )( S ′ ∩ ∩ Σ π ). § We are the result of an evolution of resolvers of the survival problem.
Argument
The resolvers hierarchy suggests an evolution of resolvers of the survivalproblem, see § ℜ [ u ] = U , see § Comment
Our species is Turing complete. Therefore we must explain the evolution ofTuring completeness. § § ¶ · The problem theory is the union of set theory and computing theory. The integrationof the two theories is achieved by using a new vocabulary to refer to old concepts, butmainly by giving the old theories a purpose that they did not have: to resolve problems.For example, a set defined by intension is named a problem, and the same set defined byextension is named its set of solutions. While both still refer to the same set, as it is thecase in set theory, the status of each of them is now very different: one is a question andthe other is an answer. And when the problem theory states that computing is resolving,it is calling a set resolvable if it is recursively enumerable, but mainly it is saying thatthe transition from intension to extension has to be calculated, because it is not writtenmagically in “The Book”; someone has to write it. ¶ · The purpose of resolving problems is not final, but the main conclusion of the paper,the Thesis § ¶ · The final Thesis § ℜ [ u ]can model everything , and then everything can be a solution, as it is stated in Theorem § everythings are not absolute, but limited to what is computable, see § § everything is everything thatis computable. This way a restriction of computing theory, countability, is inherited byproblem theory and transferred to set theory; see the details below in Subsection § § ¶ · Nevertheless, besides that main Thesis § ww.ramoncasares.com PT 42 § ¶ · In computing everything is countable, see § § § § § λ -calculus, see Curry & Feys (1958) Chapter 4. ¶ · Therefore, our way to control paradoxes in set theory, and then in this paper, is toconfine ourselves to Turing universes. But don’t worry; if this is a Turing universe, as itseems to be, then we are only excluding imaginary universes. ¶ · For example, the mathematical theorem that states that everything is a solution isproved, and it makes sense, see § P ⊂ S , but P = ( S → B ) = 2 S , and then | P | = | S | = 2 | S | > | S | , by Cantor’stheorem. It is not a paradox in a Turing universe because the forth equality is false init. The second equality is false in a Turing universe because, as we saw in Lemma § S → B ) ∗ ⊂ S . The thirdequality is true, though it follows the second one! And the forth equality is false in aTuring universe because the number of computable sets is countable, so, if | S ∗ | = ℵ ,then | S | ∗ = ℵ < ℵ = 2 | S ∗ | , that is, | S | ∗ < | S | . Therefore, P ∗ ⊂ S ∗ is true, but P ∗ is theset of computable predicates, that is, P ∗ = E of § S ] ∗ ⊂ S ∗ is also true, but[2 S ] ∗ is the set of computable sets, that is, [2 S ] ∗ = R of § S ∗ ,the set of solutions, is the set of everything that is computable. ¶ · We have just rejected the uncountable case, where | P | > ℵ , but there are two otherpossibilities: the (infinite) countable case, where | S ∗ | = | P ∗ | = | R ∗ | = ℵ , see § | S Γ | < | P Γ | < | R Γ | < ℵ , see § ¶ · We are finite, so it would be natural to restrict our investigations to the finite case,calling for finiteness instead of calling for countableness. But the finite case is trivial,and more importantly, the difference between an unrestricted universal computer anda finite universal computer is not qualitative but quantitative. There is not any stepof any calculation that an unrestricted universal computer can compute and a finiteuniversal computer cannot compute, see § § ¶ · Finally, the rejections of finiteness and uncountableness imply that countableness isthe golden mean. This is Pythagorean heaven revisited, everything is countable , but thistime we have rescued the terrifying √
2, and other irrational numbers. As Kronecker said:“God made counting numbers; all else is the work of man”. ww.ramoncasares.com
PT 43 § ¶ · Is it possible to resolve a non-computable problem? A problem is computable if,by definition, see § § § ¶ · Nevertheless you may think otherwise, and say that there is another way of resolving,let us call it ‘intuition’, that is not computable. If that were the case, then the problemtheory with its mathematical formulation, as presented in this paper, would capture theconcept of ‘computable problem’, but not the whole concept of ‘problem’. In order to seethis, please consider the following two statements: ◦ Some problems are computable. ◦ A universal computer can execute any computable resolution.Even if you believe that there are resolvable problems that are not computable, you canstill decide easily that both are true; the first is a fact, and the second is a theorem. Andthen everything in this paper would still be true of computable problems, computableresolutions, and computable solutions. ¶ · The key point in this discussion is that ‘intuition’ would refute Turing’s thesis, see § § § § § § ww.ramoncasares.com PT 44
References
Cantor (1895): Georg Cantor, “Contributions to the Founding of the Theory of Trans-finite Numbers (First Article)”; in
Mathematische Annalen , vol. xlvi, pp. 481–512,1895. Translated by P.E.B. Jourdain in
Contributions to the Founding of the Theoryof Transfinite Numbers , Dover, New York, 1955. isbn: arXiv:1209.5036 .Casares (2016b): Ram´on Casares, “Syntax Evolution: Problems and Recursion”; arXiv:1508.03040 .Curry & Feys (1958): Haskell B. Curry, and Robert Feys, with William Craig,
Combi-natory Logic , Vol. I. North-Holland Publishing Company, Amsterdam, 1958. isbn: doi:
The Kleene Symposium (editors: J. Barwise, H.J.Keisler & K. Kunen), Volume 101 of Studies in Logic and the Foundations of Math-ematics; North-Holland, Amsterdam, 1980, pp. 123–148; isbn: λ -Definability and Recursiveness”; in Duke Mathe-matical Journal , vol. 2, pp. 340–353, 1936, doi:
Proceedings ofthe IRE , vol. 47, no. 11, pp. 1940–1951, November 1959, doi:
TheJournal of Symbolic Logic , vol. 1, no. 3, pp. 103–105, September 1936, doi:
Bulletin of the American Mathematical Society , vol. 50, no. 5,pp. 284–316, 1944, doi:
Proceedings of the London Mathematical Society , vol. s2-42, issue 1, pp. 230–265, 1937, doi: λ -Definability”; in The Journal ofSymbolic Logic , vol. 2, no. 4, pp. 153–163, December 1937, doi:
Proceedings of the London Mathematical Society , vol. s2-45, issue 1, pp.161–228, 1939, doi:doi: