AAn Algebra of Properties of Binary Relations
Jochen Burghardt jochen.burghardt alumni.tu-berlin.de
Feb 2021
Abstract
We consider all 16 unary operations that, given a homogeneous binary relation R , definea new one by a boolean combination of xRy and yRx . Operations can be composed,and connected by pointwise-defined logical junctors. We consider the usual properties ofrelations, and allow them to be lifted by prepending an operation.We investigate extensional equality between lifted properties (e.g. a relation is connexiff its complement is asymmetric ), and give a table to decide this equality. Supportedby a counter-example generator and a resolution theorem prover, we investigate all 3-atom implications between lifted properties, and give a sound and complete axiom set forthem (containing e.g. “if R ’s complement is left Euclidean and R is right serial, then R ’ssymmetric kernel is left serial” ). Keywords:
Binary relation; Boolean algebra; Hypotheses generation
Contents1 Introduction 32 Definitions 53 An algebra of unary operations on relations 94 Equivalent lifted properties 17 a r X i v : . [ m a t h . L O ] F e b ist of Figures q = 5 . . . . . . . . . . . . . . . . . 114 Composition of unary operations ( p ◦ q =?) . . . . . . . . . . . . . . . . . . 115 Left inverses w.r.t. composition (? ◦ q = p ) . . . . . . . . . . . . . . . . . . 126 Right inverses w.r.t. composition ( p ◦ ? = q ) . . . . . . . . . . . . . . . . . . 137 Partition split algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 Redundant properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Computed partitions of lifted properties . . . . . . . . . . . . . . . . . . . 1910 Default representations of lifted properties . . . . . . . . . . . . . . . . . . 2011 Approach to find implications . . . . . . . . . . . . . . . . . . . . . . . . . 3012 Counter-example relation for aiir and qij0 . . . . . . . . . . . . . . . . . . . 3313 Counter-example relation for eiit . . . . . . . . . . . . . . . . . . . . . . . . 3314 Counter-example relation for g0ih . . . . . . . . . . . . . . . . . . . . . . . 3415 Counter-example relation for qbvk . . . . . . . . . . . . . . . . . . . . . . . 3416 Counter-example relation for qiiu , r0jq , and reuu . . . . . . . . . . . . . . . 3417 Counter-example relation for qism . . . . . . . . . . . . . . . . . . . . . . . 3418 Counter-example relation for qisr . . . . . . . . . . . . . . . . . . . . . . . 3519 Counter-example relation for rg0u and rgak . . . . . . . . . . . . . . . . . . 3520 Counter-example relation for rylq and rymo . . . . . . . . . . . . . . . . . . 3521 Counter-example relation for xrkv . . . . . . . . . . . . . . . . . . . . . . . 3522 Relation in Exm. 34 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3723 Relation in Exm. 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3824 Relation in Exm. 36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3825 Relation in Exm. 37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3926 Relation in Exm. 39 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3927 Relation in Exm. 40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4028 Implications about anti-symmetry (an) . . . . . . . . . . . . . . . . . . . . 4129 Implications about anti-transitivity (at) . . . . . . . . . . . . . . . . . . . . 4230 Implications about density (de) . . . . . . . . . . . . . . . . . . . . . . . . 4231 Implications about left Euclideanness (le) . . . . . . . . . . . . . . . . . . . 4332 Implications about left quasi-reflexivity (lq) . . . . . . . . . . . . . . . . . 4333 Implications about left seriality (ls) . . . . . . . . . . . . . . . . . . . . . . 4434 Implications about left uniqueness (lu) . . . . . . . . . . . . . . . . . . . . 4435 Implications about semi-order property 1 (s1) . . . . . . . . . . . . . . . . 4536 Implications about semi-order property 2 (s2) . . . . . . . . . . . . . . . . 4537 Implications about transitivity (tr) . . . . . . . . . . . . . . . . . . . . . . 4638 Implications about symmetry (sy), asymmetry (as), and reflexivity (rf) . . 4639 Implications about basic properties . . . . . . . . . . . . . . . . . . . . . . 4740 Inconsistencies between lifted properties . . . . . . . . . . . . . . . . . . . 4841 Proof sketch in Lem. 42 (lf) Order-based axiom set computation (rg) 4942 Minimal axiom set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5243 Proof tree of “Trans ∧ Irrefl → ASym” . . . . . . . . . . . . . . . . . . . . 5344 Disproof tree of “ASym ∧ Trans (cid:54)→
Refl” . . . . . . . . . . . . . . . . . . . 532 . Introduction
We strive to compute new laws about homogeneous binary relations. In [1], we consid-ered the 24 “best-known” basic properties of relations, and used a counter-example searchcombined with the Quine-McCluskey algorithm and followed by a manual confirmation /disconfirmation process, to obtain a complete list of laws of the form ∀ R. prop ( R ) ∨ . . . ∨ prop n ( R ) , where prop i are negated or unnegated properties, and R ranges over all relations. Aclassical example of such a law is “ each irreflexive and transitive relation is asymmetric ”;a lesser known example is “ a left unique and right serial relation (on a domain of (cid:62) elements) cannot be incomparability-transitive ”. Once such a law has been stated it isusually easy to prove.We also briefly sketched in [1, p.6-7] a more general approach based on regular treegrammars which could obtain laws involving arbitrary operators on relations (union, clo-sures, . . . ); however, this approach would require infeasible large computation time andmemory.In the current report, we investigate an intermediate approach which allows one to de-tect laws involving certain unary operations on relations. The considered set of operationscan be motivated by the definition of quasi-transitivity: a relation R is quasi-transitiveiff op ( R ) is transitive, where the relation op ( R ) is defined as x op ( R ) y iff x R y and not y R x. Slightly more general, we will allow all unary operations that can be defined by a booleancombination of xRy and yRx . Since there are only 16 such operations, generating lawsuggestions will be still feasible. This allows one to search for laws of the form ∀ R. prop ( op ◦ ... ◦ op ( R )) ∨ . . . ∨ prop n ( op n1 ◦ ... ◦ op nm n ( R )) , An example law is “ the symmetric closure of a left-serial and transitive relation is alwaysdense” ; once it is stated, it is straight forward to prove (Lem. 49.6).Our set of unary operations is closed w.r.t. composition (“ ◦ ”, Def. 8), therefore it issufficient to look for laws of the form ∀ R. prop ( op ( R )) ∨ . . . ∨ prop n ( op n ( R )) . We define a “lifted property” to be the composition of a basic property and a unary op-eration, that is, an expression of the form λR. prop ( op ( R )) in Lambda-calculus notation.Considering all 24 ·
16 = 384 possible combinations of basic properties (Def. 1) and unaryoperations (Def. 6), it turns out that many combinations are extensionally equal, andthere are only 81 different ones (Thm. 22).Found laws may now involve e.g. the converse, the complement, the symmetric closure,and the symmetric kernel of a relation. Moreover, some basic properties are rendered Note that we can’t express e.g. reflexive closure, since our unary operations aren’t concerned with in-formation about xRx . To achieve this, our approach could be extended to all 65536 boolean combinationsof xRy , yRx , xRx , and yRy ; however, this is likely to be beyond feasibility again. ∀ R. lprop ( R ) ∧ lprop ( R ) → lprop ( R ) , where lprop i are (unnegated) lifted properties. Using an approach different from Quine-McCluskey to eliminate redundant laws, we isolate in Sect. 5.7 a total of 124 “axioms”which imply all found laws (Thm. 56). Our C source code is provided in the ancillary files.4 . Definitions In this section, we give some preparatory definitions. Definition 1 defines the “ba-sic” properties we consider throughout this report. Definition 2 indicates their historicalorigins, and at the same time names some applications for them. Definition 3 intro-duces some operators on relations; each of them will later be identified with an admittedunary operation (Lem. 12). Definition 4 defines the notion of monotonic and antitonicproperties; Lem. 5 classifies our basic properties by these categories.
Definition 1. (Binary relation properties)
Let X be a set. A (homogeneous) binaryrelation R on X is a subset of X × X . The relation R is called1. reflexive (“Refl”, “rf”) if ∀ x ∈ X. xRx ;2. irreflexive (“Irrefl”, “ir”) if ∀ x ∈ X. ¬ xRx ;3. co-reflexive (“CoRefl”, “cr”) if ∀ x, y ∈ X. xRy → x = y ;4. left quasi-reflexive (“lq”) if ∀ x, y ∈ X. xRy → xRx ;5. right quasi-reflexive (“rq”) if ∀ x, y ∈ X. xRy → yRy ;6. quasi-reflexive (“QuasiRefl”) if it is both left and right quasi-reflexive;7. symmetric (“Sym”, “sy”) if ∀ x, y ∈ X. xRy → yRx ;8. asymmetric (“ASym”, “as”) if ∀ x, y ∈ X. xRy → ¬ yRx ;9. anti-symmetric (“AntiSym”, “an”) if ∀ x, y ∈ X. xRy ∧ x (cid:54) = y → ¬ yRx ;10. semi-connex (“SemiConnex”, “sc”) if ∀ x, y ∈ X. xRy ∨ yRx ∨ x = y ;11. connex (“Connex”, “co”) if ∀ x, y ∈ X. xRy ∨ yRx ;12. transitive (“Trans”, “tr”) if ∀ x, y, z ∈ X. xRy ∧ yRz → xRz ;13. anti-transitive (“AntiTrans”, “at”) if ∀ x, y, z ∈ X. xRy ∧ yRz → ¬ xRz ;14. quasi-transitive (“QuasiTrans”, “qt”) if ∀ x, y, z ∈ X. xRy ∧ ¬ yRx ∧ yRz ∧ ¬ zRy → xRz ∧ ¬ zRx ;15. right Euclidean (“RgEucl”, “re”) if ∀ x, y, z ∈ X. xRy ∧ xRz → yRz ;16. left Euclidean (“LfEucl”, “le”) if ∀ x, y, z ∈ X. yRx ∧ zRx → yRz ;17. semi-order property 1 (“SemiOrd1”, “s1”) if ∀ w, x, y, z ∈ X. wRx ∧ ¬ xRy ∧ ¬ yRx ∧ yRz → wRz ;18. semi-order property 2 (“SemiOrd2”, “s2”) if ∀ w, x, y, z ∈ X. xRy ∧ yRz → wRx ∨ xRw ∨ wRy ∨ yRw ∨ wRz ∨ zRw .19. right serial (“RgSerial”, “rs”) if ∀ x ∈ X ∃ y ∈ X. xRy
20. left serial (“LfSerial”, “ls”) if ∀ y ∈ X ∃ x ∈ X. xRy
21. dense (“Dense”, “de”) if ∀ x, z ∈ X ∃ y ∈ X. xRz → xRy ∧ yRz .22. incomparability-transitive (“IncTrans”, “it”) if ∀ x, y, z ∈ X. ¬ xRy ∧ ¬ yRx ∧ ¬ yRz ∧ ¬ zRy → ¬ xRz ∧ ¬ zRx .23. left unique (“LfUnique”, “lu”) if ∀ x , x , y ∈ X x Ry ∧ x Ry → x = x .24. right unique (“RgUnique”, “ru”) if ∀ x, y , y ∈ X xRy ∧ xRy → y = y .The capitalized abbreviations in parentheses are used by our software; the two-letter codesare used in tables and pictures when space is scarce.We say that x, y are incomparable w.r.t. R , if ¬ xRy ∧ ¬ yRx holds. (cid:3) Definition 2. (Kinds of binary relations)
A binary relation R on a set X is called5. an equivalence if it is reflexive, symmetric, and transitive;2. a partial equivalence if it is symmetric and transitive;3. a tolerance relation if it is reflexive and symmetric;4. idempotent if it is dense and transitive;5. trichotomous if it is irreflexive, asymmetric, and semi-connex;6. a non-strict partial order if it is reflexive, anti-symmetric, and transitive;7. a strict partial order if it is irreflexive, asymmetric, and transitive;8. a semi-order if it is asymmetric and satisfies semi-order properties 1 and 2;9. a preorder if it is reflexive and transitive;10. a weak ordering ifit is irreflexive, asymmetric, transitive, and incomparability-transitive;11. a partial function if it is right unique;12. a total function if it is right unique and right serial;13. an injective function if it is left unique, right unique, and right serial;14. a surjective function if it is right unique and and left and right serial;15. a bijective function if it is left and right unique and left and right serial. (cid:3) Definition 3. (Operation names)
1. The symmetric kernel of a relation R is defined as the largest subset of R that is asymmetric relation.2. The symmetric closure of a relation R is defined as the smallest superset of R thatis a symmetric relation.3. We define the asymmetric kernel of a relation R as the intersection of all maximalsubsets of R that are asymmetric relations.Note that for an arbitrary relation R , a largest subset that is an asymmetric relationneed not exist. For example, on the set X = { , } , the relation R = {(cid:104) , (cid:105) , (cid:104) , (cid:105)} has three asymmetric subsets, viz. R = {(cid:104) , (cid:105)} , R = {(cid:104) , (cid:105)} , and R = {} . While R and R are maximal w.r.t. ⊆ , none of them is largest. (cid:3) Definition 4. (Monotonicity)
A property prop of binary relations is called monotonicif ∀ R , R . R ⊆ R ∧ prop ( R ) ⇒ prop ( R ). It is called antitonic if ∀ R , R . R ⊇ R ∧ prop ( R ) ⇒ prop ( R ). (cid:3) Lemma 5. (Monotonic properties)
We use first-order formulas without constants andfunction symbols and with equality and one binary relation symbol R to define propertiesof binary relations, as in Def. 1. Such a formula is called an ∧ - ∨ -formula if it is in prenexnormal form, contains no other binary junctors than ( ∧ ) and ( ∨ ), and contains ( ¬ ) onlyapplied to atoms.1. A property is monotonic if its definition can be written as a closed ∧ - ∨ -formulawithout any negated occurrence of R .2. A property is antitonic if its definition can be written as a closed ∧ - ∨ -formulawithout any unnegated occurrence of R .6. The following basic properties from Def. 1 are monotonic:Refl, SemiConnex, Connex, RgSerial, LfSerial.4. The following basic properties are antitonic:Irrefl, Corefl, ASym, AntiSym, AntiTrans, LfUnique, RgUnique.5. The following basic properties are neither monotonic nor antitonic:LfQuasiRefl, RgQuasiRefl, QuasiRefl, Sym, Trans, QuasiTrans, RgEucl, LfEucl,SemiOrd1, SemiOrd2, Dense, IncTrans. Proof.
1. Given an ∧ - ∨ -formula ψ in n free variables x , . . . , x n and a relation R ona domain X , define M ( R, ψ ) to be the set of all n -tuples in X n satisfying ψ , w.r.t. R . For an ∧ - ∨ -formula ψ without quantifiers or negated occurrences of R , show byinduction on the structure of ψ that R ⊆ R implies M ( R , ψ ) ⊆ M ( R , ψ ): • If ψ has the form ψ ∧ ψ , then M ( R , ψ ) = M ( R , ψ ) ∩ M ( R , ψ ) ⊆ M ( R , ψ ) ∩ M ( R , ψ ) = M ( R , ψ ). • The case ψ ∨ ψ is similar, relying on the monotonicity of ∪ , rather than of ∩ . • If ψ has the form x i Rx j , then M ( R , ψ ) = {(cid:104) x , ..., x i , ..., x j , ..., x n (cid:105) | x i R x j } ⊆ {(cid:104) x , ..., x i , ..., x j , ..., x n (cid:105) | x i R x j } = M ( R , ψ ). • If ψ does not contain R , then M ( R , ψ ) = M ( R , ψ ).Note that ψ needn’t be an atom in this case.Finally, show that M ( R , ψ ) ⊆ M ( R , ψ ) implies both M ( R , ∀ x n .ψ ) ⊆ M ( R , ∀ x n .ψ ) and M ( R , ∃ x n .ψ ) ⊆ M ( R , ∃ x n .ψ ). By induction on n , this extendsto arbitrarily long quantifier prefixes.2. The proof is similar to 1.3. For each of the listed properties, its definition in Def. 1 is in the form required by 1.4. The definition of irreflexivity in Def. 1 is in the form required by 2. For the remainingproperties, resolving ( → ) brings it into that form: • Corefl: ∀ x, y ∈ X. ¬ xRy ∨ x = y • ASym: ∀ x, y ∈ X. ¬ xRy ∨ ¬ yRx • AntiSym: ∀ x, y ∈ X. ¬ xRy ∨ ¬ yRx ∨ x = y • AntiTrans: ∀ x, y, z ∈ X. ¬ xRy ∨ ¬ yRz ∨ ¬ xRz . • LfUnique: ∀ x , x , y ∈ X. ¬ x Ry ∨ ¬ x Ry ∨ x = x . • RgUnique: ∀ x, y , y ∈ X. ¬ xRy ∨ ¬ xRy ∨ x = x .5. For each property, we give relations R ⊂ R ⊂ R on the domain X = { , , , } such that R , but neither R nor R , has the property. R consists of all blackpairs, R consists of all black or green pairs, and R consists of all pairs. Intuitively,adding the green pair establishes the property, and adding the red destroys it again.We use semi-colons to indicate the sub-relation separations in grey-scale renderings. • LfQuasiRefl: {(cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • RgQuasiRefl, QuasiRefl: {(cid:104) , (cid:105) , (cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • Sym: {(cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • Trans, QuasiTrans: {(cid:104) , (cid:105) , (cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} RgEucl: {(cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • LfEucl: {(cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • SemiOrd1: {(cid:104) , (cid:105) , (cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • SemiOrd2: {(cid:104) , (cid:105) , (cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • Dense: {(cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} • IncTrans: {(cid:104) , (cid:105) , (cid:104) , (cid:105) ; (cid:104) , (cid:105) ; (cid:104) , (cid:105)} (cid:3) Ry : false false true true yRx : false true false true xR q y : q q q q Figure 1: Encoding of unary operations
3. An algebra of unary operations on relations
In this section, we introduce and investigate our admitted unary operations on rela-tions.
Definition 6. (Unary operations)
We represent a unary operation on relations as a 4-bitnumber p ∈ { , . . . , , A, . . . , F } , and denote its bits as p , p , p , p , that is, p = 8 · p + 4 · p + 2 · p + 1 · p . Given a binary relation R on a domain set X , and x, y ∈ X , we write R p to denote theapplication of p to R , which we define as xR p y ⇔ ( ¬ xRy ∧ ¬ yRx ∧ p ) ∨ ( ¬ xRy ∧ yRx ∧ p ) ∨ ( xRy ∧ ¬ yRx ∧ p ) ∨ ( xRy ∧ yRx ∧ p )tacitly identifying 0 with false and 1 with true , see Fig. 1. For example, xR y is true iffboth xRy and yRx is; moreover, op ( R ) from Sect. 1 can now be written as R . Figure 2shows the semantics of each of the 16 possible unary operations. They allow us to expresse.g. the converse, the complement, the symmetric kernel, and the symmetric closure of arelation. For unary operations p and q , we define p ⊆ q bitwise as p (cid:54) q ∧ p (cid:54) q ∧ p (cid:54) q ∧ p (cid:54) q . Moreover, we extend boolean connectives to unary operations in a bitwise manner, e.g. ¬ p is defined as bitwise complement:( ¬ p ) = 8 · ( ¬ p ) + 4 · ( ¬ p ) + 2 · ( ¬ p ) + 1 · ( ¬ p )= 8 · (1 − p ) + 4 · (1 − p ) + 2 · (1 − p ) + 1 · (1 − p ) , similar for the other connectives. (cid:3) It may be helpful to think of a unary operation as a set of graph rewriting rules. Thisview is supported in column “Rewriting” of Fig. 2. Representing a binary relation as adirected graph, two given vertices x and y can be connected by We use ⊕ to denote exclusive or. |||| ← → ↔ false |||| |||| |||| |||| empty relation1 0001 xRy ∧ yRx |||| |||| |||| ↔ symmetric kernel2 0010 xRy ∧ ¬ yRx |||| ← → |||| asymmetric kernel3 0011 xRy |||| ← → ↔ identity4 0100 ¬ xRy ∧ yRx |||| → ← |||| converse asymmetric kernel5 0101 yRx |||| → ← ↔ converse6 0110 xRy ⊕ yRx |||| ↔ ↔ |||| xRy ∨ yRx |||| ↔ ↔ ↔ symmetric closure8 1000 ¬ xRy ∧ ¬ yRx ↔ |||| |||| |||| incomparable9 1001 xRy ⊕ ¬ yRx ↔ |||| |||| ↔ A ¬ yRx ↔ ← → |||| complement of converse B xRy ∨ ¬ yRx ↔ ← → ↔ complement of converse asymmetric kernel C ¬ xRy ↔ → ← |||| complement D ¬ xRy ∨ yRx ↔ → ← ↔ complement of asymmetric kernel E ¬ xRy ∨ ¬ yRx ↔ ↔ ↔ |||| complement of symmetric kernel F true ↔ ↔ ↔ ↔ universal relation Figure 2: Semantics of unary operations
1. no edge at all (depicted |||| ),2. just an edge from y to x (depicted ← ),3. just an edge from x to y (depicted → ), or4. both an edge from x to y and a reverse edge (depicted ↔ ).For each of the four situations, column “Rewriting” gives the appropriate replacementperformed by an operation. For example, operation 7 replaces all ← and all → situationsby ↔ , and thus obtains the symmetric closure. Since the column “ ← ” is just a mirror of“ → ”, it is grayed out. Observe that 0 and 1 in the most significant bit in column “bin”corresponds to |||| and ↔ , respectively; similar for the least significant bit; for the twomiddle bits, 00, 01, 10, and 11 correspond to |||| |||| , ← → , → ← , and ↔ ↔ , respectively. Lemma 7. (Boolean connectives on operations) xR ¬ p y iff ¬ xR p y xR p y ∧ xR q y iff xR p ∧ q y
3. Any other boolean connective distributes over operation application in a similarway.4. If p ⊆ q , then R p ⊆ R q . Proof. • xRy ∧ yRx :Then xR ¬ p y iff ( ¬ p ) = 1 iff p = 0 iff ¬ xR p y .And xR p y ∧ xR q y iff ( p = 1) ∧ ( q = 1) iff ( p ∧ q ) = 1 iff xR p ∧ q y .10 Ry yRx xR q y yR q x x ( R q ) p y −→ q = 0 q = 0 −→ p −→ q = 1 q = 0 −→ p −→ q = 0 q = 1 −→ p −→ q = 1 q = 1 −→ p Figure 3: Composition computation example for q = 5 p \ q A B C D E F
E F
A B C D E F
C D A B E F
E F E F E F F E
F F
F F F F
F FA F E B A D C
B F F B B D D F F F F B B D D F FC F E D C B A
D F F D D B B F F F F D D B B F FE F E F E F E
F F F F F F F F F F F F F F F F F
Figure 4: Composition of unary operations ( p ◦ q =?) • xRy ∧ ¬ yRx : • ¬ xRy ∧ yRx : • ¬ xRy ∧ ¬ yRx : These cases are similar, using p , p , and p instead of p .3. Any other boolean connective c distributes over operation application, since ¬ and ∧ do, and c can be obtained as an expression over ¬ and ∧ ; e.g. xR p ∨ q y iff xR ¬ ( ¬ p ∧¬ q ) y iff ¬ ( ¬ xR p y ∧ ¬ xR q y ) iff xR p y ∨ xR q y .4. Follows from 3: p ⊆ q iff ( p ∧ q ) = p , iff ( R p ∩ R q ) = R p iff R p ⊆ R q . (cid:3) As an example, Sen’s construction [2, p.381] of a transitive relation I and a symmetricrelation P from a given quasi-transitive relation R , such that R = I . ∪ P , cf. [1, Lem.17.3,p.26], can now be paraphrased as I := R and P := R . Using Lem. 7, the proof ofdisjoint union boils down to two simple computations: I ∪ P = R ∪ R = R = R , and I ∩ P = R ∩ R = R = {} . See Lem. 52.1 for another proof using Lem. 7. Definition 8. (Operation composition)
We define operation composition in the usual wayby R p ◦ q = ( R q ) p . The set of operations is closed w.r.t. composition; Fig. 3 shows, by wayof an example ( q = 5) how to compute the bit representation of p ◦ q , given that of p and q . (cid:3) \ q A B C D E F
ACE
ACE A C A C C A C A
ACE E E ACE E E
ACE
ACE
A A C
B AB B CD D B B D DC C A
D CD D AB B D D B BE ACE E E F ABCDEF BDF EF F EF F BDF BDF BDF BDF F F F F BDF
BDF
Figure 5: Left inverses w.r.t. composition (? ◦ q = p ) Figure 4 shows a computer-generated composition table. Observe that operation 3 isa neutral element, and the operations 3 , , A, C have inverses; in fact, this set is a groupw.r.t. composition. Figure 5 shows the sets of left inverses w.r.t composition; in line p ,column q , all operations x are listed that satisfy x ◦ q = p . To save space, we omittedbraces and commas. Similarly, Fig. 6 shows the right inverses; line p , column q containsall x such that p ◦ x = q .A machine-supported investigation of the algebraic structure w.r.t. compositionshowed nothing interesting. Of the 65536 possible operation sets, 461 are closed w.r.t.composition; besides { , , A, C } the subsets { } , { F } , { , E } , { , } { , } , and { B, D } are maximal groups, each with a different neutral element. Each of the 296 closed sub-set containing 3 is, of course, a monoid; besides them, the subsets { , , E, F } , { , , } , { , , , F } , and { B, D, F } are maximal monoids, again each with a different neutral el-ement. The ancillary file optnComposition GroupsMonoids.txt lists each operation setthat is closed w.r.t. composition, and indicates whether it is a monoid, or even a group. Definition 9. (Lifted property) If prop is a property of binary relations, and q is a unaryoperation, then we call q - prop a lifted property. We define that a relation R has theproperty q - prop if R q has the property prop . (cid:3) In this way, unary operations can be lifted from relations to relation properties. Forexample, 8-transitivity is a synonym for incomparability-transitivity, and irreflexivity co-incides with C -reflexivity. 12 \ q A B C D E F ∗ AC BD E F EF AB CD A B C D E F EF CD AB C D A B E F EF ABCD
ACE BDF BDF ACE
ABCD
EFA F E B A D C B AB CD EFC F E D C B A D CD AB EFE F E BD AC F ∗ Figure 6: Right inverses w.r.t. composition ( p ◦ ? = q ) we obtain 24 ·
16 = 384 liftedproperties, some of which coincide. We call two lifted properties equivalent if they agreeon every relation on a sufficiently large, finite or infinite, domain set. For example, 3-transitivity is equivalent to 5-transitivity, due to the self-duality of the definition. Wefirst show a few simple laws about lifted properties that are needed later on. In the nextsection, we compute equivalence classes of lifted properties. Lemma 10. (Operations and properties)
1. Operation 0 always yields the empty relation.2. Operation F always yields the universal relation.3. Operations B, D, F always yield a connex relation.4. Operations 0, 2, 4, 6 always yield an irreflexive relation.5. Operations 1, 3, 5, 7 preserve reflexivity and irreflexivity.6. Operations 8, A, C, E invert reflexivity and irreflexivity.7. Operations 0, 2, 4 always yield an asymmetric relation.8. Operations 9, B, D, F always yield a reflexive relation.9. Operations 0, 9, B, D, F always yield a dense relation.10. Operations 9, B, D, F always yield a left serial relation.11. Operations 0, 9, B, D, F always yield a left quasi-reflexive relation.12. Operations 0, B, D, F always yield a relation that satisfies semi-order property 1.13. Operations 0, B, D, F always yield a relation that satisfies semi-order property 2.14. Operations 0, 1, 6, 7, 8, 9, E, F always yield a symmetric relation.
Proof.
1. Obvious from Def. 6. See [1, Exm.74, p.48] for the properties satisfied bythe empty relation,2. Obvious from Def. 6. See [1, Exm.75, p.48] for the properties satisfied by theuniversal relation,3. Each listed operation q satisfies 7 ◦ q = F , hence the symmetric closure of its resultrelation is the universal one.4. xR q x boils down to false for q ∈ { , , , } .5. For q ∈ { , , , } , the formula xR q x boils down to xRx ; hence R q is (ir)reflexiveiff R is.6. For q ∈ { , A, C, E } , the formula xR q x boils down to ¬ xRx ; hence R q is reflexiveiff R is irreflexive, and R q is irreflexive iff R is reflexive.7. Each listed operation q satisfies q ∩ (5 ◦ q ) = 0, hence its result relation is disjointfrom its own converse.8. xR x iff xRx ⊕ ¬ xRx iff true . Similarly, xR q x boils down to true for q ∈ { B, D, F } .9. The empty relation R is dense by [1, Exm.74.11, p.48]. Operations 9, B, D, Falways yield a reflexive relation by 8, which is dense by [1, Lem.48.1, p.38]. Unlike in [1], we this time included left-, right-, and two-sided quasi-reflexivity, to obtain machine-generated evidence for the laws about them; cf. Fig. 8. We didn’t include co-transitivity (which we onlyrecently became aware of) since it obviously can be expressed as C-Trans. It will turn out that two properties agree on every relation on a 7-element domain iff they agree onevery relation of a larger domain, see Thm. 22. R is quasi-reflexive by [1, Exm.74.1, p.48], hence left quasi-reflexive in particular. Operations 9, B, D, F always yield a reflexive relation by 8,which is quasi-reflexive by [1, Lem.9, p.23], and hence left quasi-reflexive.12. The empty relation R satisfies semi-order property 1 by [1, Exm.74.9, p.48]. Opera-tions B, D, F always yield a connex relation by Lem. 10.3, which satisfies semi-orderproperty 1 by [1, Lem.66, p.45].13. The empty relation R satisfies semi-order property 2 by [1, Exm.74.9, p.48]. Opera-tions B, D, F always yield a connex relation by Lem. 10.3, which satisfies semi-orderproperty 2 by [1, Lem.66, p.45].14. Each listed operation q satisfies q = 5 ◦ q , hence its result relation agrees with itsown converse. (cid:3) Lemma 11. (Unsatisfiable lifted properties)
On a domain set X with (cid:62) (cid:62) (cid:62) Proof.
On a nonempty domain set:1. By Lem. 10.8, operations 9, B, D, F yield a reflexive relation, which cannot beasymmetric by [1, Lem.10, p.23].2. By Lem. 10.8, operations 9, B, D, F yield a reflexive relation, which cannot beanti-transitive by [1, Lem.10, p.23].3. Operation 0 always yields the empty relation which isn’t left-serial on a non-emptydomain.4. By Lem. 10.4, operations 0, 2, 4, 6 yield an irreflexive relation, which cannot bereflexive on a nonempty domain.On a domain set with (cid:62) x (cid:54) = y , we have xR F y ∧ yR F x , contradicting anti-symmetry.On a domain set with (cid:62) (cid:3) emma 12. (Operation names)
1. The symmetric kernel of a relation R is obtained as R .2. The symmetric closure of a relation R is obtained as R .3. The asymmetric kernel of a relation R is obtained as R . As discussed in Def. 3.3,it is an asymmetric subset of R , but need not be a maximal one. Proof.
1. We show that R is the largest symmetric sub-relation of R . First, R issymmetric by Lem. 10.14, and a subset of R by Lem. 7.4. If R (cid:48) ⊆ R is a symmetricrelation, then xR (cid:48) y implies yR (cid:48) x by symmetry, hence xRy ∧ yRx by the subsetproperty, hence xR y by definition.2. We show that R is the smallest symmetric super-relation of R . First, R is sym-metric by Lem. 10.14, and a superset of R by Lem. 7.4. Let R (cid:48) ⊇ R be a symmetricrelation. If xR y , then by definition xRy or yRx . In the former case, we have xR (cid:48) y per superset, in the latter, we additionally use the symmetry of R (cid:48) .3. We show that R equals the intersection of all maximal asymmetric sub-relations of R .“ ⊆ ”: Let R (cid:48) be a maximal asymmetric subset of R , we show R ⊆ R (cid:48) : Let xR y ,then xRy ∧ ¬ yRx , hence ¬ yR (cid:48) x . If xR (cid:48) y did not hold, then R (cid:48) ∪ {(cid:104) x, y (cid:105)} wasa larger, but still asymmetric subset of R .“ ⊇ ”: First, {} is an asymmetric subset of R , hence a maximal one can also be found;we call it R . Now let xR (cid:48) y for every maximal asymmetric subset R (cid:48) of R , weshow xR y . We have in particular xR y , hence xRy , Assume for contradiction yRx does also hold. Then R = ( R \ {(cid:104) x, y (cid:105)} ) ∪ {(cid:104) y, x (cid:105)} is another maximalasymmetric subset of R , which, however, does not satisfy xR y , contrary toour assumption.Note that R is asymmetric by Lem. 10.7, and a subset of R by Lem. 7.4. Incontrast, R is different from the asymmetric kernel since it is not a subset of R . (cid:3) We use Zorn’s lemma here. -LfSerial3-LfUnique X X xyxRy (cid:45) (cid:12)(cid:13)(cid:24)(cid:25) (cid:27)(cid:27) truefalse
Set of partitions Current partition Current relation
Figure 7: Partition split algorithm
4. Equivalent lifted properties
In this section, we investigate extensional equality of lifted properties. We show that,starting from the basic property set from Def. 1, a of total 81 different equivalence classesexist (Thm. 22).To obtain an approximation of equivalence, we implemented a partition-refinementroutine: initially, all 384 lifted properties are in one partition; cycling though each relation R on a small finite set, we split each partition according to the property behavior on R . Figure 7 shows a snapshot of the algorithm, when it is about to split the partition { , C-LfSerial } into two singleton sets, since the current relation is not leftunique, but its complement is left serial. Running the routine on the relations over a 5-element set, we ended up with 80 partitions; for a 6-element set, the very same partitionswere obtained; checking on a 7-element set would take far too long.The raw output about the computed partitions is available in the ancillary file lpEqns.txt ; each partition is represented as an equation chain like .Most of them are easily seen to be in fact equivalences, we give the formal proofs in thefollowing. Some equivalences allow us to define one property in terms of another, seeFig. 8 and Lem. 14 below.Figure 9 shows the partitions, omitting the redundant properties for brevity. Each boxin the left part denotes one partition. The two topmost boxes denote the trivial partitions,viz. that of properties that hold for no relation at all (left) and for every relation (right);only two of the 46 and 17 members, are shown, respectively. These two partitions togethercontain all lifted properties starting with 0- or F-, along with many others, like 2-Refland 7-Sym.Partitions containing only different operations prepended to a single basic propertyare called pure partitions. They are not shown as a box, but listed in the right part ofFig. 9. For example, the first line is short notation for the raw output chains and . Singleton partitions are a subclassof pure partitions; they are listed separately. For example, the first line is short notationfor the trivial raw output chains and E-ASym . The property 0- prop applies to a relation R iff prop applies to the empty relation, independent of R .Hence, 0- prop is either true on all relations or false on all. Similar for F- prop . Note that the actual chains in file lpEqns.txt are longer than shown here, since they still contain ↔ ↔ ↔ ↔ E-ASymEmpty ↔ ↔ C-ReflConnex ↔ C-ASymSemiConnex ↔ C-AntiSymIncTrans ↔ ↔ ↔ ↔ ↔ Figure 8: Redundant properties
By construction, the computed partitioning is a coarsening of the proper extensionalequality partitioning. To obtain the latter, it sufficient to split the largest box partitionalong the dotted line, as we will show in Sect. 4.1, in particular in Thm. 22. We thus arriveat a total of 81 equivalence classes of lifted properties. We chose a default representationfor each class, they are shown in Figure 10. Considering all of them in our naive Quine-McCluskey implementation from [1, Sect.6] would require more than 2 bits, i.e. 274877millions of Tera bytes. This still renders that approach infeasible.However, it could be used to investigate small subsets of properties, looking for lawsuggestions including complement, converse, etc. For example, to investigate the connec-tions between quasi-transitivity and semi-orders, one may restrict oneself to QuasiTrans,SemiOrd1, SemiOrd2, and ASym. From Fig. 10, it can seen that 21 distinct propertieswould be needed. However, we didn’t yet perform such an investigation.Instead, we checked all possible implications between default representations up to alength of 3, i.e. with at most 2 antecedents. This is presented in Sect. 5. redundant properties. The figure shows, for each unary operation q and each basic property prop , the default representationto which q - prop is extensionally equal. If q - prop is its own default representation, it is highlighted in blueand cyan, for a non-singleton and a singleton equivalence class, respectively. The trivial partitions aredenoted by + and − . +, -, as, 7-as, C-as, E-as, s1, 1-s1, 2-s1, 7-s1, C-s1, s2, 2-s2, C-s2, sy, 1-tr, 2-tr, 7-tr, 8-tr, 9-tr, E-tr. Pure Partitions
ASym: 1 ↔ ↔
5, 8 ↔ A ↔ C AntiSym: 1 ↔ ↔
5, 8 ↔ A ↔ C Refl: 1 ↔ ↔ ↔
7, 8 ↔ A ↔ C ↔ E SemiOrd1: 1 ↔ E , 2 ↔
4, 3 ↔
5, 7 ↔ A ↔ C SemiOrd2: 2 ↔
4, 3 ↔ A ↔ C AntiTrans: 2 ↔
4, 3 ↔ A ↔ C Trans: 2 ↔
4, 3 ↔ B ↔ D , A ↔ C Dense: 3 ↔ A ↔ C Singleton Partitions
ASym: 7, EAntiSym: 7, EAntiTrans: 1, 6, 7, 8, EDense: 1, 6, 7, 8, ELfEucl: 3, 5, A, CLfSerial: 1, 2, 3, 4, 5, 6, 7, 8, A, C, ELfUnique: 1, 2, 3, 4, 5, 6, 7, 8, A, C, ELfQuasiRefl: 1, 3, 5, 7, 8, A, C, E
Figure 9: Computed partitions of lifted properties
19 1 2 3 4 5 6 7 8 9 A B C D E Fem Empty + as sy 7-as sy 7-as sy 7-as C-as − E-as − E-as − E-as − un Univ − E-as − E-as − E-as − C-as 7-as sy 7-as sy 7-as sy 3-as +rf Refl − rf − rf − rf − rf C-rf + C-rf + C-rf + C-rf +Irrefl + C-rf + C-rf + C-rf + C-rf rf − rf − rf − rf − CoRefl + an sy 7-an sy 7-an sy 7-an C-an 9-an E-an − E-an − E-an − lq LfQuasiRefl + 1-lq sy lq sy 5-lq sy 7-lq 8-lq + A-lq + C-lq + E-lq +RgQuasiRefl + 1-lq sy 5-lq sy lq sy 7-lq 8-lq + C-lq + A-lq + E-lq +QuasiRefl + 1-lq sy 7-lq sy 7-lq sy 7-lq 8-lq + E-lq + E-lq + E-lq +sy Sym + + sy sy sy sy + + + + sy sy sy sy + +as ASym + as + as + as sy 7-as C-as − C-as − C-as − E-as − an AntiSym + an + an + an sy 7-an C-an 9-an C-an 9-an C-an 9-an E-an − SemiConnex − E-an 9-an C-an 9-an C-an 9-an C-an 7-an sy an + an + an +Connex − E-as − C-as − C-as − C-as 7-as sy as + as + as +tr Trans + 1-tr 2-tr tr 2-tr tr sy 7-tr 8-tr 9-tr C-tr D-tr C-tr D-tr E-tr +at AntiTrans + 1-at 2-at at 2-at at 6-at 7-at 8-at − C-at − C-at − E-at − QuasiTrans + + 2-tr 2-tr 2-tr 2-tr + + + + 2-tr 2-tr 2-tr 2-tr + +RgEucl + 1-tr sy 5-le sy le sy 7-tr 8-tr 9-tr C-le sy A-le sy E-tr +le LfEucl + 1-tr sy le sy 5-le sy 7-tr 8-tr 9-tr A-le sy C-le sy E-tr +s1 SemiOrd1 + 1-s1 2-s1 s1 2-s1 s1 sy 7-s1 7-s1 sy C-s1 + C-s1 + 1-s1 +s2 SemiOrd2 + E-tr 2-s2 s2 2-s2 s2 9-tr 8-tr 7-tr sy C-s2 + C-s2 + 1-tr +RgSerial − − − C-lu − E-lu − RgUnique + 1-lu 4-lu 5-lu 2-lu lu 6-lu 7-lu 8-lu 9-an C-lu − A-lu − E-lu − Figure 10: Default representations of lifted properties .1. Proof of equivalence classes Lemma 13.
1. If R is symmetric, then R = R = R = R are the empty relation,and R = R B = R D = R F are the universal relation.2. If R is asymmetric, then R = R are the empty relation, and R E = R F are theuniversal relation. Proof.
1. Symmetry implies R = R , hence R = R (2 ◦ = R , similar for the otheroperations.2. Asymmetry implies R = R , hence R = R (1 ◦ = R , similar for E . (cid:3) Lemma 14. (Redundant properties)
1. RgSerial ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ E-ASym7. Empty ⇔ ⇔ C-Refl9. Connex ⇔ C-ASym10. SemiConnex ⇔ C-AntiSym11. IncTrans ⇔ ⇔ ⇔ Proof. xRx iff xR x by definition.“ ⇒ ”: If xR y , then xRy ∨ yRx ; both cases imply xRx by quasi-reflexivity.“ ⇐ ”: If xRy , then xRy ∨ yRx by weakening, hence xR y , and similarly yR x , hence xR x and yR y .6. “ ⇒ ”: Let xR E y , then ¬ xRy ∨ ¬ yRx ; both cases contradict universality.“ ⇐ ”: Let R E be asymmetric; it is also symmetric by Lem. 10.14, hence empty by[1, Lem.16, p.25], hence ¬ xRy ∨ ¬ yRx never holds, hence R is universal.7. “ ⇒ ”: xR y would imply xRy ∨ yRx , contradicting R ’s emptiness.“ ⇐ ”: xRy would imply xRy ∨ yRx by weakening, hence xR y and yR x , contra-dicting asymmetry.8. Since ¬ xRx iff xR C x , for all x .9. “ ⇒ ”: Let xR C y , then ¬ xRy by definition, hence yRx by connexity, hence ¬ yR C x .“ ⇐ ”: If ¬ xRy , then xR C y , hence ¬ yR C x by asymmetry, hence yRx .210. Similar to 9:“ ⇒ ”: Let xR C y , then ¬ xRy by definition, then x = y or hence yRx by semi-connexity, hence x = y or ¬ yR C x .“ ⇐ ”: If ¬ xRy , then xR C y , hence x = y or ¬ yR C x by anti-symmetry, hence x = y or yRx .11. By definition.12. “ ⇒ ”: If xR y and yR x , then xRy ∨ yRx ; both cases imply x = y by co-reflexivity.“ ⇐ ”: Let xRy , then xRy ∨ yRx by weakening; this implies both xR y and yR x ,hence x = y by anti-symmetry.13. By definition. (cid:3) Lemma 15.
1. If R is transitive, then its complement R C satisfies semi-order prop-erty 2.2. If R is symmetric and satisfies semi-order property 2, then its complement R C istransitive. Proof.
1. Let xR C y and yR C z hold, assume for contradiction that neither wR C x nor xR C w nor wR C y nor yR C w nor wR C z nor zR C w holds. From xRw and wRy , weobtain xRy by transitivity, contradicting our assumption.2. Let xR C y and yR C z hold, assume for contradiction xRz . Then zRx by symmetry,and xRy ∨ yRx ∨ yRz ∨ zRy by semi-order property 2. Due to symmetry, the latterdisjunction boils down to xRy ∨ yRz , contradicting our assumption. (cid:3) Lemma 16.
For q ∈ { , , , , , , E, F } , a relation R is q -left-Euclidean iff it is q -transitive. Proof.
By Lem. 10.14, R q is symmetric. Hence q -left-Euclideanness and q -transitivitycoincide by [1, Lem.36, p.33]. (cid:3) Lemma 17. (Characterization of symmetry)
All of the following properties are equiva-lent:1. 6-AntiSym2. 6-ASym3. 2-, 4-, 6-LfQuasiRefl4. 2-, 4-, 6-, B-, D-LfEucl5. 6-, 9-SemiOrd16. 9-SemiOrd27. 2-, 3-, 4-, 5-, A-, B-, C-, D-Sym8. 6-Trans 22 roof.
We show for each property that is applies to a relation R iff R is symmetric.For the “if” part, observe that for a symmetric R , we have R = R = R = R emptyby Lem. 13.1, and hence asymmetric and anti-symmetric by [1, Exm.74.6, p.48], leftEuclidean by [1, Exm.74.2, p.48], quasi-reflexive by [1, Exm.74.1, p.48], symmetric by [1,Exm.74.3, p.48], transitive by [1, Exm.74.7, p.48], and satisfy semi-order property 1 and 2by [1, Exm.74.9, p.48]. Moreover, for a symmetric R , we have that R = R B = R D = R F are universal by Lem. 13.1, and hence left Euclidean by [1, Exm.75.1, p.48], symmetric by[1, Exm.75.2, p.48] and satisfy semi-order property 1 and 2 by [1, Exm.75.5, p.48]. Andfor symmetric R we have R symmetric by self-duality, R C symmetric by contraposition,and R A symmetric by duality to R C .For the “only if” part:1. Let R be anti-symmetric, we show that R is symmetric. Let xRy , assume forcontradiction ¬ yRx ; then x (cid:54) = y . Moreover, xR y by definition, hence ¬ yR x byanti-symmetry, contradicting Lem. 10.14.2. Follows from 1 and [1, Lem.13.2, p.25].3. • Case 2: Let xRy , assume for contradiction ¬ yRx . Then xR y , hence xR x byleft quasi-reflexivity; this contradicts Lem. 10.4. • Cases 4 and 6 are shown similar.4. • Cases 2, 4, and 6 follow from 3, using [1, Lem.46, p.37]. • Case B : Let xRy , then xR B y by definition. Moreover yR B y by Lem. 10.8,hence yR B x by Euclideaness, hence yRx ∨ ¬ xRy by definition, hence yRx byour assumption. • Case D is similar.5. • Case 6: Let xRy , assume for contradiction ¬ yRx . Then xR y by defini-tion, hence yR x by Lem. 10.14. Moreover, ¬ xR x by Lem. 10.8. Applyingsemi-order property 1 to yR x , ¬ xR x , and xR y infers yR y , contradictingLem. 10.4. • Case 9: Let xRy , assume for contradiction ¬ yRx . Then ¬ xR y and also ¬ yR x by definition. Moreover, xR x and yR y by Lem. 10.8. Applying semi-order property 1 to xR x , x, y incomparable w.r.t. R , and yR y infers xR y ,contradicting the incomparability of x, y .6. By Lem. 10.8, R is reflexive, hence also connex by [1, Lem.66, p.45]. Since R isalso symmetric by Lem. 10.14, it is universal by [1, Lem.53.2, p.40]. By definition,this means that xRy and yRx agree everywhere, i.e. that R is symmetric.7. • Case 2: Let xRy , assume for contradiction ¬ yRx . Then xR y by definition,hence yR by symmetry, hence yRx by definition, contradicting our assump-tion. • Case B : Let xRy , then xR B y by definition, hence yR B x by symmetry, hence yRx by definition, since ¬ xRy cannot hold. • Cases 4 and D follow by duality. • Case 3 is trivial. • Cases 5, A , and C follow by duality and contraposition.8. Let xRy , assume for contradiction ¬ yRx . Then, xR y and also yR x by definition,hence xR x by transitivity, contradicting Lem. 10.4. (cid:3) emma 18. (Symmetry and semi-order property 1) If R is symmetric and satisfies semi-order property 1, then R C does, too. Proof.
First, we have R symmetric iff xRy ↔ yRx iff ¬ xRy ↔ ¬ yRx iff R C symmetric.Now let R be symmetric and satisfy semi-order property 1, let wR C x , ¬ xR C y , ¬ yR C x ,and yR C z hold. Assume for contradiction ¬ wR C z . Then by definition xRy , ¬ yRz , and wRz , hence by symmetry ¬ zRy , and zRw . Applying semi-order property 1 to these factsimplies xRw , hence wRx , contradicting wR C x . (cid:3) Lemma 19.
1. A relation is 2-dense iff it is 4-dense.2. If R is symmetric, then R is dense.3. The converse direction does not hold if the universe set has (cid:62) Proof. R is the converse relation of R , since 4 = 5 ◦
2. Since the definition ofdensity is self-dual, we are done.2. By Lem. 13.1, R is empty, hence dense by [1, Exm.74.11, p.48].3. [1, Exm.76, p.49] and Fig. 12 here show an example relation on a 7-element domainthat is 2-dense and not symmetric. (cid:3) Lemma 20.
The following are equivalent:1. R is anti-symmetric and semi-connex.2. R is the identity relation.3. R is left-unique.4. R is anti-symmetric.5. R B is anti-symmetric6. R D is anti-symmetric. Proof.
By Lem. 10.14 and Lem. 10.8, R is always symmetric and reflexive. • ⇒
2: Let xR y hold, by definition, we have two cases: – If xRy ∧ yRx , then x = y by anti-symmetry of R . – If ¬ xRy ∧ ¬ yRx , then x = y since R is semi-connex.This shows that R is co-reflexive. Since it is also reflexive, it must be the identity,by [1, Lem.5.1, p.21]. • ⇒ – Anti-symmetry: Let xRy ∧ yRx , then xR y by definition, hence x = y . – Semi-connex: Let ¬ xRy ∧ ¬ yRx , then again xR y by definition, hence x = y . • ⇒ ∧
4: trivial. 24 ⇒ R is anti-symmetric, then it is co-reflexive by [1, Lem.7.7, p.22], hence the identityby [1, Lem.5.1, p.21]. • ⇒ R is co-reflexive by [1, Lem.7.1, p.22], hence the identity by [1, Lem.5.1,p.21]. • ∨ ⇒ R B and R D are supersets of R , by Lem. 5.4, anti-symmetry isantitonic. • ⇒ ∧
6: Let xR B y and yR B x , then by definition xRy ∨ ¬ yRx and yRx ∨ ¬ xRy .This is logically equivalent to ( xRy ∧ yRx ) ∨ ( ¬ xRy ∧ ¬ yRx ), that is, to xR y . Since R is the identity as shown above, we have x = y . The proof for 6 is dual. (cid:3) Lemma 21. (Derived equivalences)
Let prop , prop i denote properties of binary relations.1. If ∀ R. prop ( R p ) → prop ( R p ), then ∀ R. prop ( R p ◦ r ) → prop ( R p ◦ r ) for everyoperation r .2. If 3- prop ↔ prop , then 2- prop ↔ prop , A- prop ↔ C- prop , and B- prop ↔ D- prop .This applies to the properties Refl, QuasiRefl, Sym, ASym, AntiSym, Trans, Anti-Trans, SemiOrd1, SemiOrd2, and Dense.3. If 1- prop ↔ prop , then 0- prop ↔ prop , 1- prop ↔ prop , 8- prop ↔ A- prop , 9- prop ↔ B- prop , and 8- prop ↔ C- prop .This applies to the properties Refl, ASym, and AntiSym.4. If 3- prop ↔ prop , then 2- prop ↔ prop , 4- prop ↔ prop , 5- prop ↔ prop , A- prop ↔ E- prop , B- prop ↔ F- prop , C- prop ↔ E- prop , and D- prop ↔ F- prop .This applies to the properties Refl, and QuasiRefl.5. If 1- prop ↔ E- prop , then 0- prop ↔ F- prop , 6- prop ↔ prop , and 7- prop ↔ prop .This applies to the properties Sym, and SemiOrd1. Proof.
1. Since R p i ◦ r = ( R r ) p i , the consequent is an instance of the antecedent.2. Apply 1 to r = 2 , A, B . The listed properties are self-dual by Def. 1.3. Apply 1 to r = 4 , , A, B, C . To prove the list: • If R is reflexive, then R is, too, by Lem. 10.5. If R is reflexive, then itssuperset R is, too, by Lem. 5.3. • If R is asymmetric, then its subset R is, too, by Lem. 5.4. If R is asymmetric,it is empty by [1, Lem.16, p.25] using Lem. 10.14, that is, xRy ∧ yRx can neverhappen, that is, R is asymmetric. • If R is anti-symmetric, then its subset R is, too, by Lem. 5.4. If R is isanti-symmetric, then it is co-reflexive by [1, Lem.7.7, p.22] using Lem. 10.14,that is, xRy ∧ yRx can happen only for x = y , that is, R is anti-symmetric.4. Apply 1 to r = 2 , , , A, B, C, D . To prove the list: • If R is reflexive, then R is, too, by Lem. 10.5. If R is reflexive, then xRx ∨ xRx for all x , hence R is reflexive. 25 Let R be quasi-reflexive, let xR y hold. Then xRy ∨ yRx by definition. Eachalternative implies xRx ∧ yRy , and hence xR x ∧ yR y .Conversely, let R be quasi-reflexive, let xRy hold. Then xR y , hence xR x ∧ yR y , which boils down to xRx ∧ yRy .5. Apply 1 to r = 0 , ,
7. To prove the list: • Both R and R E are always symmetric by Lem. 10.14. • For semi-order property 1, the equivalence follows from Lem. 18, since R E =( R ) C and R = ( R E ) C . (cid:3) Theorem 22. (Equivalent lifted properties)
On a set with (cid:62) { , } must be split off from the partition of 3-Sym (indicated by thedotted line). Proof.
First, 2-Dense is not equivalent to 3-Sym: for the usual < relation on the rationalnumbers, 3-Dense and 2-Dense coincide since < is asymmetric; therefore < satisfies 2-Dense, but not 3-Sym; see also [1, Exm.76, p.49]. This shows the split is necessary.By construction of the partition split algorithm from Fig. 7, no two lifted propertiesfrom different partitions in Fig. 9 can agree on all relations. Therefore, it is sufficient toshow that each two members of a partition are equivalent, with the above exception. • Boxes:All properties in the topmost left and right box are equivalent by Lem. 11 and 10,respectively. The two members of the split-off class are equivalent by Lem. 19.1; allmembers of the remainder of that partition are equivalent by Lem. 17.The bottommost 5 right boxes are covered by Lem. 15 (using symmetry of R , R , R , R , R E by Lem. 10.14) and 16.The right box second from top is covered by Lem. 20. • Pure partitions:The ASym and the AntiSym partitions are covered by Lem. 21.3; the Refl partitionsby this and Lem. 21.4.The SemiOrd1 partition are covered by Lem. 21.2 and Lem. 21.5; the SemiOrd2,AntiTrans, Trans, and Dense partitions by Lem. 21.2.Note that not all equivalences implied by Lem. 21 lead to pure partitions, e.g. B-Dense ↔ D-Dense holds trivially (by Lem. 10.8 and [1, 48.1, p.38]) and is thereforereflected in the universal (“+”) partition. • Singleton Partitions:Nothing to show. (cid:3)
Lemma 23. (Default representations)
26. For every “+” in the matrix in Fig. 10, the operation of its column always yields arelation satisfying the property of its row. For example, operation 9 always yields areflexive and symmetric relation.2. For every “ − ”, the column operation never yields a relation satisfying the rowproperty. For example, operation B never yields a anti-transitive or left-uniquerelation.3. Whenever in a row prop the default representations coincide for column p and q ,and q = p ◦ r , then r preserves p - prop . For example, operations 2, 3, 4, 5, A, B, C,D all preserve symmetry.This generalizes Lem. 10. Proof.
By Thm. 22, two properties are equivalent iff Fig. 10 shows the same defaultrepresentation for them.1. If operation p - prop has default representation +, then R p always satisfies prop .2. If operation p - prop has default representation − , then R p never satisfies prop .3. Let R satisfy p - prop , then R equivalently satisfies q - prop . That is, ( R r ) p = R q satisfies prop , hence R r satisfies p - prop . (cid:3) Lemma 24.
A relation R is E-AntiSym iff its complement is CoRefl. Proof. R E is AntiSym iff R ◦ C is AntiSym iff R C is 7-AntiSym iff R C is CoRefl. (cid:3) . Implications between lifted properties The approach of Sect. 4 could produce only “equivalence laws”, of the form ∀ R. prop ( op ( R )) ↔ prop ( op ( R )) . On the other hand, investigating all “disjunction laws”, of the form ∀ R. prop ( op ( R )) ∨ . . . ∨ prop n ( op n ( R )) , for prop i a negated or unnegated property, is computationally infeasible for large n , asdiscussed in Sect. 4. As a compromise, we investigate in this section “3-implication laws”,of the form ∀ R. prop ( op ( R )) ∧ prop ( op ( R )) → prop ( op ( R )) , that is, implications of unnegated lifted properties with up to 2 antecedents and oneconclusion. Note that we don’t cover the negation of a lifted property; in particular, itcannot be obtained by composing it with the C operation; for example, “ R is not reflexive”is usually different from “ R is C -reflexive, a.k.a. irreflexive”. However, we can express true and false as 0-Sym and 0-Refl, respectively; hence we can express incompatibilitybetween two lifted properties.Since the Quine-McCluskey procedure would require too much computational re-sources, we used a different approach. We cycled though all 81 = 531441 triples (cid:104) op - prop , op - prop , op - prop (cid:105) of default representation of lifted properties, and for eachof them searched a counter-example relation R such that prop ( op ( R )) ∧ prop ( op ( R )) ∧¬ prop ( op ( R )) holds. We recorded the search result in an array, indexed by the rankof the three default representations. In case no counter-example was found, we askedan external resolution prover to prove prop ( op ( R )) ∧ prop ( op ( R )) → prop ( op ( R )),which might, or might not, succeed. In the former case, we recorded this result in the ar-ray. The remaining array cells were handled manually, unless they were trivial (Sect. 5.2). To reference a 3-implication under consideration, we use its array indices, encodedin base 27, with digits represented as , a , b , c , . . . , z . For example, the implication“SemiOrd1 ∧ Refl → Connex” corresponds to array indices [18][14][11] , that is, toarray cell (18 ·
81 + 14) ·
81 + 11 = 119243 = ((6 ·
27 + 1) ·
27 + 15) ·
27 + 11, and istherefore encoded as faok . This encoding ensures an implication is referenced by a fixedcode, unique across multiple program runs, and independent of the order in which laws In [1], 58 of the found “disjunction laws” (of basic properties only) had (cid:62) The latter presupposes a nonempty relation domain. The equivalence classes of lifted properties shown in Fig. 9 are numbered in the following order, cf.the list lpnfList[] in file lpImplicationsTables.c : boxed partitions starting at rank 0: 0-rf, 0-sy,3-sy, 2-de, 9-an, 1-tr, 7-tr, 8-tr, 9-tr, E-tr; pure partitions starting at rank 10, top to bottom, left to right:3-as, C-as, 3-an, C-an, . . . , C-de; singleton partitions starting at rank 33, in the same order: 7-as, E-as,7-an, E-an, . . . , E-lq. EProver version
E 2.3 Gielle , from resolveImplCode.sh can be used to convert base-27 codesinto implications; the call “ nonprominentProperties -lpImpl resolve ” can be usedfor the reverse direction.
We implemented routines to draw trivial conclusions from a found negative (counter-example) or positive (proof) result. An inference is considered trivial if it doesn’t requireany knowledge about basic property definitions (as given in Def. 1), except for monotonic-ity information as provided in Lem. 5.3 and 5.4.The trivial inference mechanism is intended to reduce the load on both thecounter-example search and the external theorem prover. For example, when “LfEucl → LfQuasiRefl” is already established, we needn’t attempt to prove “RgEucl → RgQuasiRefl” since this follows trivially by taking the converse relation. When a trivialproof or disproof of an implication is already known, we run neither the counter-examplesearch nor the external theorem prover for it. All trivial proofs are printed to a log fileduring a program run; that output may subsequently be used to extract the trivial part ofa proof tree of a given implication, by running program justify . As an example, Fig. 43shows the proof tree for “Trans ∧ Irrefl → ASym” (base-27 code ijrj ); it also demonstratesthat trivial proofs needn’t be intuitively obvious. Figure 44 shows the disproof tree of“ASym ∧ Trans (cid:54)→
Refl” ( clcn ); right to each base-27 code, the corresponding index tripleis shown.
Definition 25. (Trivial inference rules)
Using
I, J, K, I ( · ) , . . . to denote default repre-sentations of lifted properties, our trivial inference rules are:1. if I ∧ J → K , then J ∧ I → K ;2. if J ∧ I (cid:54)→ K , then I ∧ J (cid:54)→ K ;3. if I ∧ J → K , and H ∧ I → J , then H ∧ I → K , and variants thereof;4. if H ∧ I (cid:54)→ K , and H ∧ I → J , then I ∧ J (cid:54)→ K , and variants thereof;5. if ∀ R. I ( R ) ∧ J ( R ) → K ( R ), then ∀ R, p. I ( R p ) ∧ J ( R p ) → K ( R p );6. if ∃ R, p. I ( R p ) ∧ J ( R p ) (cid:54)→ K ( R p ) then ∃ R. I ( R ) ∧ J ( R ) (cid:54)→ K ( R ). (cid:3) The above example trivial inference from “LfEucl → LfQuasiRefl” to “RgEucl → RgQuasiRefl” is achieved by applying rule 25.5 with op = 5.In the last two rules, it does make sense to consider all lifted properties equivalentto a given one during the search for a common operation op . For example, from theimplication C-tr ∧ → C-as ( isok ), we may infer A-tr ∧ → ∧ → op = A , which normalizes to the defaultrepresentation 3-tr ∧ C-rf → ijrj ); cf. the two topmost lines in the proof tree inFig. 43.Initially, we fill all array cells corresponding to trivial implications, including mono-tonicity information: Definition 26. (Trivial initialization rules)
Our trivial initialization rules are:1. x ∧ y → true ,2. x ∧ false → z ,3. false ∧ y → z , 29 (cid:26)(cid:24)(cid:25)(cid:26)(cid:25) in (cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:122) (cid:27)(cid:26)(cid:24)(cid:25)(cid:26)(cid:25) out (cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:24)(cid:58) [00][00][00] ... [80][80][80] ++val wg [00][00][80][10][28][14][14][14][76][28][15][18] + − +? (cid:27) init (cid:45) trivial inferences (cid:12)(cid:13) (cid:27) (cid:45) counter-example generator (cid:12)(cid:13) (cid:27) (cid:45) EProver (cid:12)(cid:13) (cid:27)(cid:27) (cid:106) (cid:0)(cid:0) (cid:64)(cid:64) manualproofs
Figure 11: Approach to find implications x ∧ y → x (cid:48) if x → x (cid:48) is known by monotonicity or antitonicity, and5. x ∧ y → y (cid:48) if y → y (cid:48) is known, similarly. (cid:3) For example, since ASym is antitonic, we initialize array cell [33][3][10] , corre-sponding to “7-ASym ∧ → k0ij ), to true , by rule 26.4.It turned out that no other than the above 6 inference rules are needed. For example,we don’t need an extra “explosion” rule: if I ∧ J → false is known to hold, we can infer I ∧ J → K for an arbitrary lifted property K using an appropriate variant of rule 25.3and the trivial implication J ∧ false → K from 26.2. Figure 11 shows an overview of our architecture for searching for 3-implications. Thearray mentioned above is our central data structure. Various actions on it can be trig-gered by command-line options. We first perform the trivial initializations, then usuallyload results from previous runs, infer all possible trivial conclusions, and perform a dis-proof/proof search for still undecided array cells. The latter action cycles through allundecided array cells, and runs the counter-example generator and (in case of not findinga counter-example) the external prover; once a decision has been found, trivial conclusionsare drawn from it.During the early runs, when most cells were yet undecided, we drew from each newinformation as much conclusions as possible, in order to prune generator and prover calls.Since this amounted to a depth-first search, it often lead to extremely long proofs. In In Fig. 43 and 44 length-minimized proof trees of ijrj and clcn are shown. The corresponding depth-first versions have 13148 and 124989 nodes, even when their 3083 and 34310 repeated subtrees are countedas one node each, respectively. Using a save/ load mechanism, we could reuse results from earlier runs. Consistency over varyingimplementations is ensured as the notions of counter-example and external proof remainedunchanged.We kept the log files of all external proof attempts. If an implication couldn’t beproven true immediately, we retried it with increased minimum domain cardinality, upto 8. A domain cardinality of (cid:62) n was expressed by a first-order formula introducing n constants c , . . . , c n , and requiring their pairwise distinctness; for n = 1 we used theformula ¬∀ x. false . In Fig. 42, we noted any nontrivial minimum domain cardinality.Whenever the external prover noted that an implication I ∧ J → K could be provenwithout using its conclusion K (EProver outcome “contradictory axioms”), we validatedthat by manually calling the prover for I ∧ J → false . Altogether, we collected log filesof successful proof attempts of 4422 implications of the latter form, and 34727 otherimplications.A few implications that couldn’t be proven this way needed to be proven manually,viz. g0rw , uivk , uklu , uydr , voye , voyr , wfyp , see Sect. 5.8. All remaining implications wereinvalid; we provided counter examples on finite (Sect. 5.4) or countably infinite (Sect. 5.5)domains manually. Knowing about the validity / invalidity of every 3-implication is thebasis to achieve completeness of our axiom set in Sect. 5.7 below. Example 27. (Finite counter-examples)
The following implications are not universallyvalid:1. ( aiir ) 9-AntiSym ∧ (cid:54)→ qij0 ) 2-LfSerial ∧ (cid:54)→ eiit ) 1-SemiOrd1 ∧ (cid:54)→ C-SemiOrd14. ( g0ih ) 2-SemiOrd2 ∧ (cid:54)→ jijo ) 3-Dense ∧ (cid:54)→ qbvk ) 1-LfSerial ∧ (cid:54)→ qiiu ) 2-LfSerial ∧ (cid:54)→ r0jq ) 4-LfSerial ∧ (cid:54)→ reuu ) 4-LfSerial ∧ (cid:54)→ qism ) 2-LfSerial ∧ (cid:54)→ qisr ) 2-LfSerial ∧ (cid:54)→ rg0u ) 4-LfSerial ∧ (cid:54)→ rgak ) 4-LfSerial ∧ (cid:54)→ For example, we originally recorded an implication as “presumably true” when no counter-examplewas found and no external proof was available. This turned out to be tedious to implement, and inferencesfrom presumably true implications weren’t of great use; so we dropped this kind of truth value. Originally, we implemented saving and loading the complete array in binary format. However, weabandoned this approach in favor of a print / scan mechanism for lists of base-27 codes, since such filesare easier to maintain. rylq ) 6-LfSerial ∧ (cid:54)→ rymo ) 6-LfSerial ∧ (cid:54)→ xrkv ) 3-LfQuasiRefl ∧ (cid:54)→ Proof.
We show in Fig. 12 to 21 a counter-example relation on a finite domain for eachof the implications. In each picture, an arrow from x (light blunt end) to y (dark peakedend) indicates xRy , colors have only didactic purpose. All properties have been machine-checked by our implementation. Additionally, we give a witness for each dissatisfiedimplication conclusion:1. In Fig. 12, we have bRa , a incomparable to itself, aRd , but not bRd . Hence, R isnot SemiOrd1.Note that this relation is the same as in [1, Exm.76, p.49]; variations of this relationappear in the counter-examples for qij0 , eiit , g0ih , qiiu , r0jq , reuu , and xrkv .2. The same elements in this relation demonstrate that R is not Trans, and hence notQuasiTrans, since R coincides with R .3. In Fig. 13, we have bR C m , mR n , and nR C c , but not bR C c .4. In Fig. 14, we have aR h and hR b , but not aR b .5. The counter-example relation is obtained from Fig. 13 by removing the loops mRm and nRn . We then have no intermediate element for mR n .6. In Fig. 15, we have aR b and bR c , but aR c .7. In Fig. 16, we have bR a and aR c , while h is incomparable to each of a, b, c .8. In the same figure, R is not 7-Dense, since aR n has no intermediate element.9. In the same figure, R is not 2-SemiOrd2, since bR a and aR c , but h is incomparableto all of them.10. In Fig. 17, we have xR y , and yR z , but xR z .11. In Fig. 18, xR y has no intermediate element w.r.t. R : each element is comparablew.r.t. R to x or to y .12. In Fig. 19, we have x R y and y R z , but x is comparable with none of them.13. In the same figure, we have x R y and y R z , but x R z , i.e. R is not anti-transitive.14. In Fig. 20, aR b , b, c incomparable w.r.t. R , and dR d , but not aR d .15. In the same figure, bR c has no intermediate element.16. In Fig. 21, we have hR c , but not hR h . (cid:3) The shown counter-examples might not be the simplest possible. igure 12: Counter-example relation for aiir and qij0 Figure 13: Counter-example relation for eiit igure 14: Counter-example relation for g0ih Figure 15: Counter-example relation for qbvk
Figure 16: Counter-example relation for qiiu , r0jq , and reuu Figure 17: Counter-example relation for qism igure 18: Counter-example relation for qisr Figure 19: Counter-example relation for rg0u and rgak
Figure 20: Counter-example relation for rylq and rymo
Figure 21: Counter-example relation for xrkv .5. Infinite counter-examples Example 28. (Non-negative rational numbers)
The set QQ + = { q ∈ QQ | q (cid:62) } of non-negative rational numbers with the usual order ( < ) can be used to disprove the followingimplications:1. ( a0mz ) 4-Dense ∧ (cid:54)→ acib ) 2-Dense ∧ D-Trans (cid:54)→ afay ) 2-Dense ∧ (cid:54)→ Proof.
We define R = ( < ) to establish the usual notation. Since R is asymmetric, R i coincides with R i +1 , for i = 0 , . . . ,
7. We therefore have R = R = ( < ), R = ( > ), R = (=), and R D = ( (cid:62) ). It is well known that ( < ) and ( > ) are dense, ( > ) is left-serial,(=) is trivially anti-symmetric, and ( (cid:62) ) is transitive. However, ( < ) is neither left-serial(consider x = 0) nor symmetric. (cid:3) Example 29. (Integer numbers)
The set ZZ of integer numbers with the usual order ( < )can be used to disprove the following implications:1. ( jeuc ) D-Trans ∧ (cid:54)→ jeux ) D-Trans ∧ (cid:54)→ Proof.
As in Exm. 28, we write R for ( < ), and observe R = ( < ) and R D = ( (cid:62) ).The former is left-serial, and the latter is transitive. However, ( < ) is neither dense noranti-transitive. (cid:3) Example 30. (aaje)
Proof.
Consider QQ ∪ {∞} with R being the usual order ( (cid:54) ) on QQ , extended by xR ∞ for all x ∈ QQ ∪ {∞} \ { } . We have that R is anti-symmetric, since it is so on QQ , and ∞ Rx doesn’t hold, except for x = ∞ . R is ( < ) ∪ {(cid:104) x, ∞(cid:105) | (cid:54) = x (cid:54) = ∞} , which is dense:observe that QQ \ {− } (cid:51) xR ∞ has e.g. x + 1 as intermediate element, similar for x = − R C is not dense, since 0 R C ∞ has no intermediate element, as xR C ∞ implies x = 0,but neither 0 R C ∞ R C ∞ holds. (cid:3) Example 31. (aaxu)
Proof.
Consider QQ ∪ {∞} with R being the usual order ( < ), and with ∞ incomparableto all elements. Since this ordering is asymmetric, R coincides with R . Hence, it is dense,since it is dense on QQ , and ∞ is not involved in any relation pair. Moreover, R satisfiessemi-order property 1, since xR y ∧ yRz can hold only if x = y (cid:54) = ∞ ; in this case, wRx implies wRz by transitivity. But R doesn’t satisfy semi-order property 2, since 0 < < ∞ is incomparable to all of them. (cid:3) Example 32. (i0iq) · · (cid:45) (cid:117) (cid:45) (cid:117) (cid:45) (cid:117) (cid:45) (cid:117) (cid:45) (cid:117) (cid:117) (cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:88)(cid:122) Figure 22: Relation in Exm. 34
Proof.
Consider { a, b } × QQ with R defined by (cid:104) x, y (cid:105) R (cid:104) u, v (cid:105) iff x = u and y < v , em-ploying the usual order < . This results in two independent copies of the rational numbers QQ . Since this ordering is asymmetric, 2-Trans coincides with 3-Trans; both properties arewell-known to be satisfied by < . Since no elements of different copies of QQ are related, R itself is also both 2- and 3-Transitive, and moreover 2-Dense. However, R doesn’t satisfysemi-order property 1, since (cid:104) a, (cid:105) R (cid:104) a, (cid:105) , and (cid:104) a, (cid:105) is incomparable w.r.t. r to (cid:104) b, (cid:105) ,and (cid:104) b, (cid:105) R (cid:104) b, (cid:105) , but not (cid:104) a, (cid:105) R (cid:104) b, (cid:105) . (cid:3) Example 33. (jevo, jew0)
D-Trans and 2-LfSerial implies neither 1-Dense ( jevo ) nor4-LfSerial ( jew0 ). Proof.
Consider the set IN with xRy defined as x > y or x = 0 ∧ y = 1. Then R isleft serial, since e.g. x + 1 R x for x > R
0. The relation R D can be obtained as( R ) ∪ ( R ); with xR y iff x < y or x = 1 ∧ y = 0, and xR y iff x = y ; hence xR D y iff x (cid:54) y or x = 1 ∧ y = 0. Therefore, R D is transitive: Let xR D y and yR D z . • If x (cid:54) y and y (cid:54) z , then x (cid:54) z , hence xR D z . • If x (cid:54) y and y = 1 ∧ z = 0, then x = 0 or x = 1, in both cases, xR D • If x = 1 ∧ y = 0 and y (cid:54) z and z = 0, then 1 R D • If x = 1 ∧ y = 0 and y (cid:54) z and z >
0, then 1 (cid:54) z , hence xR D z . • The case x = 1 ∧ y = 0 and y = 1 ∧ z = 0 is a contradiction.However, R = {(cid:104) , (cid:105) , (cid:104) , (cid:105)} is not dense. Moreover, R is not left serial, i.e. R is notright serial, since 0 R x applies to no x . (cid:3) Example 34. (qjxu)
Proof.
Consider the universe set IN and its usual order ( > ), define R = ( > ) \{(cid:104) , (cid:105) , (cid:104) , (cid:105) , (cid:104) , (cid:105)} , cf. Fig. 22. Since ( > ), and therefore R , is asymmetric, it coincideswith R . The latter is left serial since e.g. i + 1 R i for each i (cid:62)
1, and 4 R R also satisfies semi-order property 1: Let wR x , x, y incomparable w.r.t. R , and yR z . • If x = y and z > w > x = y > z >
0, i.e. 0 is not involved, hence wR z by transitivity of ( > ). • If x = y and z = 0, then w > x = y (cid:62) > z , hence again wR z by transitivity.37 · · (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:27) (cid:90)(cid:90)(cid:90)(cid:90)(cid:125)(cid:90)(cid:90)(cid:90)(cid:90)(cid:126)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:105)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:80)(cid:113) · · · (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:27) (cid:26)(cid:26)(cid:26)(cid:26)(cid:61)(cid:26)(cid:26)(cid:26)(cid:26)(cid:62)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:41)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:16)(cid:49) · · · Figure 23: Relation in Exm. 35 · · · (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45)(cid:17)(cid:17)(cid:17)(cid:17)(cid:51) (cid:117) (cid:104) b, (cid:105)· · · (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45)(cid:81)(cid:81)(cid:81)(cid:81)(cid:115) (cid:117) (cid:104) a, (cid:105) Figure 24: Relation in Exm. 36 • The case x ∈ { , , } and y = 0 is impossible due to yR z . • If x = 0 and y ∈ { , , } ,then yR z implies 3 (cid:62) y > z >
0, and wR w (cid:62)
4, hence wR z .However, R doesn’t satisfy semi-order property 2, since 0 is incomparable to all elementsinvolved in 3 R R (cid:3) Example 35. (qlas)
Proof.
Consider the universe set { a, b } × IN , and the relation R defined by (cid:104) x, y (cid:105) R (cid:104) u, v (cid:105) iff x = u ∧ y > v or y = 0 ∧ v (cid:54) = 0 or y (cid:54) = 0 ∧ v = 0. Figure 23 sketches the relation: itconsists of two copies of ( > ) on IN , but additionally relates each zero to each non-zerouniverse member, and vice versa. We have (cid:104) x, y (cid:105) R (cid:104) u, v (cid:105) iff x = u ∧ y > v >
0; thisrelation is ls and tr. R is not E-de since (cid:104) a, (cid:105) R E (cid:104) b, (cid:105) has no intermediate element byconstruction. (cid:3) Example 36. (qldr)
Proof.
On the set { a, b } × IN , define R (cid:48) such that for all x, y, u :1. (cid:104) x, y +1 (cid:105) R (cid:48) (cid:104) x, y (cid:105) ,2. (cid:104) x, (cid:105) R (cid:48) (cid:104) u, (cid:105) ,3. (cid:104) x, (cid:105) R (cid:48) (cid:104) x, (cid:105) ,4. R (cid:48) applies to no other pairs than given above.Figure 24 sketches that relation; it is left serial due to rule 1. Let R be the transitiveclosure of R (cid:48) , then R is transitive by construction. R is obtained from R by removingthe pairs (cid:104)(cid:104) a, (cid:105) , (cid:104) a, (cid:105)(cid:105) and (cid:104)(cid:104) b, (cid:105) , (cid:104) b, (cid:105)(cid:105) ; it is left serial, since R (cid:48) is. However, R is notdense: (cid:104) a, (cid:105) R (cid:104) b, (cid:105) has no intermediate element, since (cid:104) a, (cid:105) is comparable w.r.t. R toeach element except (cid:104) b, (cid:105) , and similar for (cid:104) b, (cid:105) . (cid:3) · · (cid:45) (cid:117) (cid:104) c, (cid:105) (cid:45) (cid:117) (cid:104) c, (cid:105) (cid:45) (cid:117) (cid:104) c, (cid:105) (cid:45) (cid:117) (cid:104) c, (cid:105) (cid:45) (cid:117) (cid:104) c, (cid:105) (cid:45) (cid:117) (cid:104) c, (cid:105)· · · (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105)· · · (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:17)(cid:17)(cid:17)(cid:17)(cid:51)(cid:81)(cid:81)(cid:81)(cid:81)(cid:115)(cid:17)(cid:17)(cid:17)(cid:17)(cid:43) (cid:81)(cid:81)(cid:81)(cid:81)(cid:107) Figure 25: Relation in Exm. 37 (cid:45)(cid:64)(cid:64)(cid:64)(cid:82) (cid:45)(cid:45)(cid:72)(cid:72)(cid:72)(cid:106)(cid:72)(cid:72)(cid:72)(cid:106) (cid:45)(cid:88)(cid:88)(cid:88)(cid:122) . . . Figure 26: Relation in Exm. 39
Example 37. (qmkh)
Proof.
Consider the universe set { a, b, c } × IN , and the relation R defined as follows: (cid:104) x, y (cid:105) R (cid:104) u, v (cid:105) iff x = u ∧ y = v + 1 or y = v = 0 ∧ x (cid:54) = u ∧ b ∈ { x, u } . The universeconsists of three copies of IN , connected only at 0; see Fig. 25. The relation R is 2-LfSerial, since (cid:104) x, i + 1 (cid:105) R (cid:104) x, i (cid:105) , but not (cid:104) x, i (cid:105) R (cid:104) x, i + 1 (cid:105) for each x ∈ { a, b, c } and i ∈ IN .It is 7-AntiTrans, since it doesn’t contain an undirected cycle of length 3. In fact, theonly cycles are (cid:104) b, (cid:105) R (cid:104) a, (cid:105) R (cid:104) b, (cid:105) and (cid:104) b, (cid:105) R (cid:104) c, (cid:105) R (cid:104) b, (cid:105) , and compositions thereof, allof which have an even length. However, it is not 1-LfUnique, since (cid:104) b, (cid:105) R (cid:104) a, (cid:105) and (cid:104) b, (cid:105) R (cid:104) c, (cid:105) . (cid:3) Example 38. (qpkw)
Proof.
Consider the set IN with xRy defined as x = y + 1 or x = y = 0. Then R isleft serial, since x + 1 R x for all x . Moreover, R is right-unique, since a successor number y + 1 is related only to y , and 0 is only related to 0. However, R is not left quasi-reflexive,since e.g. 0 R
2, but not 0 R (cid:3) Example 39. (ufbk)
Proof.
Consider IN \ { } and the relation xRy : ⇔ x = y//
2, where // denotes truncatinginteger division, see Fig. 26. It is 2-LfUnique, since x = y// x implies x = x . It is4-LfSerial, since for a given y , choosing x = y · x (cid:54) = y// y = x//
2, thatis, xR y . However, it is not 4-LfUnique, since e.g. 1 = 2 // //
2, that is, 2 R R (cid:3) Example 40. (uo0e) · · (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) (cid:45) (cid:117) (cid:104) b, (cid:105) · · · (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:45) (cid:117) (cid:104) a, (cid:105) (cid:54)(cid:63) Figure 27: Relation in Exm. 40
Proof.
On the set { a, b } × IN , define R such that:1. (cid:104) x, y (cid:105) R (cid:104) x, y +1 (cid:105) ,2. (cid:104) x, (cid:105) R (cid:104) u, (cid:105) if x (cid:54) = u ,3. R applies to no other pairs than given above.Figure 27 sketches R and allows us to verify that R is left-unique (no two arrows end atthe same point) and R is left serial (a one-sided arrow starts from each point). However, R is not transitive, since (cid:104) a, (cid:105) R (cid:104) b, (cid:105) , and (cid:104) b, (cid:105) R (cid:104) a, (cid:105) , but not (cid:104) a, (cid:105) R (cid:104) a, (cid:105) . (cid:3) After several attempts, we eventually achieved to decide for every 3-implication itstruth value. We found 156384 valid and 375057 invalid implications; the former include42657 from trivial initialization.In the ancillary file lpImpl.log each implication can be found together with its truthvalue and a justification. The executable justify can be used to obtain the truth values(and justifications) of given implications. Figure 43 and 44 show examples; note thatwe don’t print proof parts contributed by the external prover or the counter-examplegenerator.Implications with one antecedent can be visualized as a directed graph (Hasse dia-gram), with lifted properties as vertices and implications as edges. This has been donefor basic properties in [1, Fig.18, p.20]. For lifted properties, the graph is still too largeto be depicted in one piece. We have split it by basic property and show the pieces inFig. 28 to 38. In each figure, vertices and edges for the depicted basic property are shownin blue, while others, for implied or implying properties, are shown in red. We have twoexceptions: edges contained in our minimal axiom set (Sect. 5.7 below) are colored green,and edges following from monotonicity or antitonicity are colored cyan (for the depictedbasic property) or magenta (for others). For brevity, the two-character codes from Def. 1are used, and the separating hyphen between operation and property name is omitted.For example, the bottommost vertex in Fig. 28 represents the lifted property 9-AntiSym.Since the vertices represent equivalence classes, all implications are strict.Figure 39 summarizes the part about all basic properties; co-transitivity is included(as “Ctr”) although we didn’t define it as a basic property. In this figure, in order toobtain a feasible complexity, a red vertex is shown only if it is both implied by a bluevertex and implying another one. in our approach, they appear as 3-implications of the form e.g. I ∧ I → K igure 28: Implications about anti-symmetry (an) Figure 40 shows the inconsistencies between lifted properties. Clusters of incompat-ibilities are presented using disjunctions of lifted properties. For example, the topmosthorizontal red edge, from “Alu,Clu,Cat” to “1at,1lu”, indicates that each property fromthe set { A-LfUnique , C-LfUnique , C-AntiTrans } is inconsistent with each property from { , } . Nested boxes indicate implications between properties, thatis, subset relations between their extensions. For example, the “as” box is inside the “1at”box since each ASym relation is 1-AntiTrans ( cjdj ). The red lines’ end points should beconsidered carefully; for example, the topmost vertical line connects “1ls” with “as” butnot with the surrounding “1at” box, since 1-LfSerial is inconsistent with ASym, but notnecessarily with 1-AntiTrans. All found inconsistencies are covered in Fig. 40; however,some of them require application of some implications, viz.6-at ⇒ lvfx ),E-an ⇒ C-an ⇔ sc ⇒ ld0m , djno ),E-at ⇒ C-at ( mvoz ),cr ⇔ ⇒ an ⇒ kuxl , dakh ),cr ⇔ ⇒ sy ( kuxb ),cr ⇔ ⇒ kuzn ),E-an ⇒ E-lu ( ldbr ), andE-an ⇒ sy ( ld0b ).For example, the inconsistency of SemiConnex and CoRefl known from [1, Lem.8.8, p.23]is tacitly understood as a consequence of the inconsistency of 1-LfUnique and 7-LfUniqueshown at the bottom of Fig. 40.Together, Fig. 28 to 40 show a refinement of the graph from [1, Fig.18, p.20]. i.e. 3-implications of the form I ∧ J → igure 29: Implications about anti-transitivity (at)Figure 30: Implications about density (de) igure 31: Implications about left Euclideanness (le)Figure 32: Implications about left quasi-reflexivity (lq) igure 33: Implications about left seriality (ls)Figure 34: Implications about left uniqueness (lu) igure 35: Implications about semi-order property 1 (s1)Figure 36: Implications about semi-order property 2 (s2) igure 37: Implications about transitivity (tr)Figure 38: Implications about symmetry (sy), asymmetry (as), and reflexivity (rf) igure 39: Implications about basic properties (cid:81)(cid:81)(cid:81) (cid:17)(cid:17)(cid:17) (cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:5)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69)(cid:69) Figure 40: Inconsistencies between lifted properties (cid:0)(cid:0)(cid:0)(cid:9) (cid:64)(cid:64)(cid:64)(cid:82)(cid:1)(cid:1)(cid:1)(cid:11) (cid:65)(cid:65)(cid:65)(cid:85) (cid:1)(cid:1)(cid:1)(cid:11) (cid:65)(cid:65)(cid:65)(cid:85) k i (cid:15)(cid:14) (cid:12)(cid:13) K \ { k } (cid:0)(cid:0)(cid:0)(cid:9) (cid:64)(cid:64)(cid:64)(cid:82)(cid:1)(cid:1)(cid:1)(cid:11) (cid:65)(cid:65)(cid:65)(cid:85) (cid:1)(cid:1)(cid:1)(cid:11) (cid:65)(cid:65)(cid:65)(cid:85) (cid:15)(cid:14) (cid:12)(cid:13) A (cid:114) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:18) (cid:98)(cid:114) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98)(cid:114) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98) (cid:0)(cid:0)(cid:0)(cid:18) (cid:98)(cid:114) (cid:114) (cid:54) (cid:98) (cid:54) (cid:98) (cid:114) (cid:54) (cid:98) (cid:54) (cid:98) (cid:114) (cid:54) (cid:98) (cid:114) (cid:54) (cid:98) (cid:114)(cid:114)(cid:114) (cid:45) (cid:98) (cid:45) (cid:98)(cid:114) (cid:45) (cid:98) (cid:45) (cid:98)(cid:114) (cid:45) (cid:98) (cid:45) (cid:98) (cid:114)(cid:114) Figure 41: Proof sketch in Lem. 42 (lf) Order-based axiom set computation (rg)
In order to obtain a human-comprehensible result, we searched for a small set of“axioms” from which all valid 3-implications could be derived by trivial inferences in thesense of Sect. 5.2. Our approach is described in an abstract way by Def. 41 and Lem. 42.
Definition 41. (Abstract derivation)
Let U (“universe”) be a set of “abstract proposi-tions”. An abstract inference rule r on U is a partial mapping from finite subsets of U to U . If r ( X ) is defined for a finite X ⊆ U , we say that r ( X ) is directly derivable from X .We say that p is derivable from a set Y if an abstract derivation tree exists with p atits root and members of Y as its leaves such that each non-leaf node corresponds to avalid direct derivation. The set Y needn’t be finite, and not each of its members need toappear as a leaf. We will assume a fixed given set of inference rules in the following.Moreover, let a fixed well-founded total order < on U be given. We say a derivationtree is ordered if every of its nodes is larger than all its (direct) descendants; we call itsroot orderly derivable from the set of its leaves. (cid:3) Lemma 42. (Axiomatization)
Let V (“valid”) be a set of propositions; let K (“kernel”)be a subset of V such that each member of V can be derived from K . Let A (“axioms”)be the set of propositions k ∈ K such that k is not orderly derivable from K \ { k } . Then:1. A is a subset of K ;2. Each member of V is derivable already from A ;3. No member a of A is orderly derivable from A \ { a } . Proof.
We first show, by induction on ( < ), that each k ∈ K is orderly derivable from A : If k is not orderly derivable from K \ { k } , the k ∈ A by construction, and we are done(trivial one-node derivation tree). Else, let k be orderly derivable from { k , . . . , k n } ⊆ K \ { k } (red tree in the left part of Fig. 41). Then k i < k , and hence by I.H. k i is orderlyderivable from A (green tree). Composing the tree to derive k from { k , . . . , k n } with thetrees for all k i gives an ordered derivation tree for k from A .Now we show the claimed properties:1. By construction of A . 49. If v ∈ V , then v has a derivation tree with leaves in K by assumption. Eachof its leaves has an (even ordered) derivation tree from A . Composing the treesappropriately yields a derivation tree for v from A (note that it needn’t be ordered).3. If a ∈ A was orderly derivable from A \ { a } , then it was also orderly derivable fromthe superset K \ { a } , contradicting the construction of A .Note, however, that some a ∈ A may still be unorderly derivable from larger mem-bers of A . (cid:3) Figure 41 (rg) shows a geometrical analogy of this approach of order-based axiom setcomputation. A member of the set V is represented by 3 adjacent circles in different colors(red, green, blue). Each color corresponds to an own proof ordering. A colored arrowindicates an ordered derivation; for simplicity we assume each inference rule to work on asingleton set. A full circle indicates an axiom w.r.t. the proof order of its color; a hollowcircle represents a non-axiom. The proof order shown in red, green, and blue, needs 3,6, and 6 axioms, respectively. Note that the set of red axioms can be further reduced byapplying, in turn, e.g. the green order to it.In our implementation, an abstract proposition consists of a valid or invalid 3-implication, like “C-Refl ∧ → ∧ (cid:54)→ eccj and ilcb , respectively). We represented such a proposition by its base-27 code, usingdifferent permutations of character positions. For example, the implications “3-Trans ∧ C-Refl → ∧ → C-ASym” correspond to ijrj and faok ,respectively; so a derivation of the former from the latter is unorderly w.r.t. the identitypermutation, but orderly when comparing the reverse codes ( jrji and koaf ). Applying xor masks to the characters before comparing, we obtain more possible orderings. Inorder to obtain best-comprehensible axioms, we prepended symbol encoding the “humansimplicity” of an implication. We used an ad hoc evolutionary algorithm based on Lem. 42 to minimize the car-dinality of the axiom set as far as we could. Starting from the set of all 34727 provenimplications, we applied Lem. 42 in parallel for a couple of different orderings. One ofthem resulted in a set of 360 axioms, the minimum number we could achieve. Repeatingthat procedure 5 times, we obtained axiom sets comprising 285, 265, 256, 253, and 252axioms. After that we manually tried to omit axioms one by one, and ended up with aset of 124 axioms.Subsequently, we manually exchanged a few axioms by equivalent ones for presentationreasons. For example, we replaced “Refl → LfQuasiRefl” ( dsqt ) by “Refl → QuasiRefl”( dsqv ), to get a nicer Hasse diagram in Fig. 39; in the presence of the remaining 123 axioms,both are equivalent. Similarly, we replaced right by left properties where possible, e.g.“RgEucl → QuasiTrans” ( pej0 ) by the equivalent “LfEucl → QuasiTrans” ( owg0 ).The result is shown in Fig. 42; each valid 3-implication can be inferred from theseaxioms using the trivial inference rules from Sect. 5.2 (Thm. 56).In Fig. 42, we marked implications where all atoms share the same basic property.We used “ ∗ ” when this can be obtained by applying just the redundancies from Fig. 8,and “ + ” when equivalences from Fig. 9 are needed in addition; for example, the top leftimplication “IncTrans → SemiOrd2” can be rephrased as “7-SemiOrd2 → SemiOrd2”. For example, the list “Refl ∧ LfEucl → → → RgSerial”, “Refl → LfSerial” is ordered from worst to best simplicity. wfzb , can be falsified on a 2-element set, but is validfor all relation domains of (cid:62) I ∧ J → K is marked “=” if also K → I ∧ J holds; it is marked “ (cid:46) ” if also K → J holds,and “ > ” if also I ∧ K → J holds. The axioms could be manually tuned such that thecases J ∧ K → I and K → I don’t occur. Note that the reverse or semi-reverse versionsare not part of the axiomatization; for example, the bottom right axiom “ASym ∧ Dense → cllc ) is marked ” > ” since ASym and 2-Dense also implies Dense ( cijd ), butthe latter can be proved from axioms adsd , cixg , cpbj , dcaa , owht , pgrm , and xzgd .Figure 42 shows the 20 “2-implication laws” first, then the single “2-inconsistency law” hpr0 , then the 103 proper “3-implication laws”. We additionally grouped implications bythe basic/lifted distinctions of their atoms.As for invalid implications, we minimized the set of implications obtained from finite(Sect. 5.4) and infinite (Sect. 5.5) counter-examples by manual ad-hoc removals; we didn’tapply Lem. 42 for them, and we didn’t try to minimize the set of computer-generatedcounter-examples. 17 implications were falsified by examples on an infinite domain, 133on a 5-element domain (computer-generated), and 16 on another finite domain. All axioms from Fig. 42 could be proven by EProver, except g0rw , uivk , uklu , uydr , voye , voyr , and wfyp . In this section, we give manual proofs for these and a few otheraxioms. Moreover, we include references to all proofs of axioms that were already givenin [1]. Lemma 43. (2-Implications group 0)
1. ( biuv ) IncTrans → SemiOrd22. ( dsqv ) Refl → ild0 ) Trans → QuasiTrans4. ( owg0 ) LfEucl → QuasiTrans5. ( owht ) LfEucl → LfQuasiRefl6. ( xzgd ) LfQuasiRefl → Dense7. ( yqnt ) QuasiRefl → LfQuasiRefl
Proof.
1. See [1, Lem.34, p.31].2. Since Refl implies QuasiRefl by [1, Lem.9, p.23]; the latter is equivalent to 7-LfQuasiRefl by Lem. 14.5.3. See [1, Lem.18, p.27].4. See [1, Lem.40, p.34].5. See [1, Lem.46, p.37].6. See [1, Lem.48.3, p.38].7. By Def. 1.6. (cid:3)
Lemma 44. (2-Implications group 1)
1. ( dspx ) Refl → iuv + it → s2 dsqv rf → qr ild0 ∗ tr → qt owht le → lq owg0 le → qt xzgd lq → de yqnt ∗ qr → lq dspx rf → fbao s1 → fbar s1 → gknw s2 → ilce ∗ tr → jlmq ∗ de → owgk le → xzhs ∗ lq → fkdd → de lmco → ir pwog + C-le → it vpm0 → qt fkdo → hpr0 → ¬ at > sy tr → le d0gh + = an sy → cr dcaa (cid:46) an qt → tr faok = s1 rf → co hlmf = at de → em ikuv tr at → s2 oskj > le an → lu phfb > re lq → sy ujry (cid:46) lu ir → at ukaa lu s1 → tr ulmt (cid:46) lu de → le xtlr lq s2 → s1 xxcn = lq rs → rf xzkv ∗ = lq rq → qr + > sy s1 → + > sy tr → (cid:46) sy ls → (cid:46) sy lu → bldc ∗ it tr → D-tr cllc > as de → cnyy (cid:46) as ls → djid ∗ = sc an → ik0q tr s1 → ikvl (cid:46) tr at → ot0q le s1 → qtaq ls s1 → qudq ls tr → uivk lu it → uklu lu s2 → uodr lu rs → uoed lu rs → xt0s lq s1 → (cid:46) sy 1-at → at (cid:46) sy 7-de → de (cid:46) sy 7-ls → ls (cid:46) sy 1-lu → lu + (cid:46) sy 1-lq → lq cpbj > as 2-lu → lu cixg > as 9-tr → it ckry (cid:46) as 2-at → at ikcr tr 7-s1 → s1 otcr > le 7-s1 → s1 pgrm (cid:46) re 8-lu → sc ulfl lu C-tr → an umxl lu 7-de → an unrb lu 1-ls → sy unuj lu 2-ls → as xtg0 lq C-s1 → qt xxjz (cid:46) lq 7-ls → ls xzog (cid:46) lq 8-lq → it ykcr > qr 7-s1 → s1 bioh ∗ it 1-tr → cmyp > as 7-de → cokb (cid:46) as 7-ls → cpnn (cid:46) as 6-lu → ioas tr 4-ls → E-de jljo de D-tr → otfp > le C-s1 → owae le 8-de → C-de uiyr lu 9-tr → ukfp lu C-s1 → unae lu 8-de → C-de uoge lu 6-ls → C-de vhqx (cid:46) ru 8-lq → A-lq ykfq qr C-s1 → btlg + (cid:46) → it eqgt (cid:46) → le npkj > → lu ntuj → as pudt > C-le tr → le qbad → de vtlj > → as vxcb → sy gtlu ∗ C-s2 s2 → ituh C-tr at → itui C-tr at → E-tr j0vb ∗ (cid:46) D-tr it → C-tr ltvl ∗ = → pxdx (cid:46) C-le rs → uydr → wcch → wxdx (cid:46) C-lu rs → adsd ∗ → de etfr ∗ > → s1 jbcr D-tr 7-s1 → s1 pkcr > A-le 7-s1 → s1 pxjz (cid:46) C-le 7-ls → ls qavd → de qbgd → de rwpd → de vxfj → as wgba → rs xlfr → s1 ewvq → fjvk → g0rw + → C-s2 nnas ∗ → E-de voye → C-de voyr → vozd → wfxh → wfyp → wfzb → Figure 42: Minimal axiom set igure 43: Proof tree of “Trans ∧ Irrefl → ASym”Figure 44: Disproof tree of “ASym ∧ Trans (cid:54)→
Refl”
2. ( fbao ) SemiOrd1 → fbar ) SemiOrd1 → gknw ) SemiOrd2 → jlmq ) Dense → owgk ) LfEucl → xzhs ) LfQuasiRefl → Proof.
1. Refl implies LfSerial and RgSerial by [1, Lem.54, p.40]; the latter conjunc-tion is equivalent to 1-LfSerial by Lem. 7.2 and 14.1.2. Let xR z , that is, xRz ∧ zRx . If xRx , then xR x , and we can choose x as inter-mediate element. If ¬ xRx , then zRz by semi-order property 1 applied to zRx , x, x incomparable, xRz ; so we can choose z as intermediate element.3. Let xR z , that is, ¬ xRz ∧ ¬ zRx , we distinguish three cases: • If ¬ xRx , then xR x , so x can be used as intermediate element. • if ¬ zRz , then z can be used, in a similar way. • If xRx ∧ zRz , then applying semi-order property 1 yields xRz , contradicting xR z .4. Let xR y , assume for contradiction ¬ xR x . The latter means xRx by definition.Hence x, y must be comparable w.r.t. R , by semi-order-property 2 applied to xRx and xRx . But xR y means that x, y are incomparable w.r.t. R .5. Let xR z , then xRz or zRx by definition. In the first case, we have xRy ∧ yRz forsome y , by density of R , hence xR y ∧ yR z by Lem. 7.4. The second case is similar.6. Let xR y and yR z , we show ¬ xR z . By definition, we have to distinguish thefollowing cases: • xRy ∧ ¬ yRx and yRz ∧ ¬ zRy :Then yRy by left Euclideanness, hence yRx by the same property, contradictingthe case assumption. 53 xRy ∧ ¬ yRx and ¬ yRz ∧ zRy :Then xRz and zRx by left Euclideanness, that is, ¬ xR z . • ¬ xRy ∧ yRx and yRz ∧ ¬ zRy :Then xRz would imply the contradiction xRy , and zRx would imply the con-tradiction zRy . But ¬ xRz ∧ ¬ zRx implies ¬ xR z . • ¬ xRy ∧ yRx and ¬ yRz ∧ zRy :Then zRx would imply the contradiction yRz , and xRz would imply zRx (since zRz ) which just has been show to contradict. Again, ¬ xRz ∧ ¬ zRx implies ¬ xR z .7. Let xR y , then xRy ∧ yRx by definition, hence xRx by left quasi-reflexivity, hence xR x by definition. (cid:3) Lemma 45. (2-Implications group 2)
1. ( fkdd ) 7-SemiOrd1 → Dense2. ( lmco ) 1-AntiTrans → C-Refl
Proof.
1. Let xRz , we distinguish two cases: • If xRx , we can use x as intermediate element. • If ¬ xRx , then applying semi-order property 1 to zR x , x, x incomparable w.r.t. R , xR z yields zR z , that is, zRz , and we can use z as intermediate element.2. Assume for contradiction ¬ xR C x for some x . Then xRx , hence xR x ∧ xR x bydefinition, hence ¬ xR x by anti-transitivity. (cid:3) Lemma 46. (2-Implications group 3)
1. ( fkdo ) 7-SemiOrd1 → Proof.
1. The proof is similar to that of 45.1: Let xR z , that is, xRz ∧ zRx , then xR z and zR x . We distinguish two cases: • If xRx , then xR x , and we can use x as intermediate element. • If ¬ xRx , then applying semi-order property 1 to zR x , x, x incomparable w.r.t. R , xR z yields zR z , that is, zRz , that is zR z , and we can use z as interme-diate element. (cid:3) Lemma 47. (2-Contradictions group 0)
1. ( hpr0 ) No relation on a set of (cid:62)
Proof.
1. Assume for contradiction that R is AntiTrans and R is LfUnique. By [1,Lem.22, p.28], AntiTrans implies Irrefl. Let w, x, y, z be 4 distinct elements. By8-LfUnique, since each element is incomparable to itself, it must be comparable tothe other three. Hence, in the directed graph corresponding to R , the vertice of w must have two incoming edges or two outgoing ones, leading w.l.o.g. to x and y such that, moreover, w.l.o.g. xRy . In the incoming case, we have xRy and yRw ,but xRw . In the outgoing case, we have wRx and xRy , but wRy . Both contradictAntiTrans. (cid:3) emma 48. (3-Implications group 0)
1. ( ) Sym ∧ Trans → LfEucl2. ( d0gh ) AntiSym ∧ Sym → CoRefl3. ( dcaa ) AntiSym ∧ QuasiTrans → Trans4. ( faok ) SemiOrd1 ∧ Refl → Connex5. ( hlmf ) AntiTrans ∧ Dense → Empty6. ( ikuv ) Trans ∧ AntiTrans → SemiOrd27. ( oskj ) LfEucl ∧ AntiSym → LfUnique8. ( phfb ) RgEucl ∧ LfQuasiRefl → Sym9. ( ujry ) LfUnique ∧ Irrefl → AntiTrans10. ( ukaa ) LfUnique ∧ SemiOrd1 → Trans11. ( ulmt ) LfUnique ∧ Dense → LfEucl12. ( xtlr ) LfQuasiRefl ∧ SemiOrd2 → SemiOrd113. ( xxcn ) LfQuasiRefl ∧ RgSerial → Refl14. ( xzkv ) LfQuasiRefl ∧ RgQuasiRefl → QuasiRefl
Proof.
1. See [1, Lem.36, p.33].2. See [1, Lem.7.7, p.22].3. See [1, Lem.19, p.27].4. See [1, Lem.66, p.45].5. See [1, Lem.49, p.39].6. See [1, Lem.24, p.28].7. See [1, Lem.45, p.37].8. See [1, Lem.37, p.33].9. Assume for contradiction xRy and yRz , but xRz . then x = y by LfUnique, contra-dicting Irrefl.10. See [1, Lem.62.2, p.43].11. See [1, Lem.47.1, p.37].12. See [1, Lem.73, p.47].13. See [1, Lem.55.2, p.40].14. By Def. 1.6. (cid:3) In Lem. 48.6, note that a relation R is both Trans and AntiTrans iff R is vacuouslytransitive, i.e. ¬∃ x, y, z. xRy ∧ yRz . Such a relation also is ASym, 1-, 2-, 6-, 7-AntiTrans,1-, 8-, C-, E-Dense, 1-, 8-, A-, C-, E-LfQuasiRefl, 8-, A-, C-, E-LfSerial, 1-LfUnique, 1-,C-SemiOrd1, 2-, 3-, C-SemiOrd2, and 1-, 2-, E-Trans.As a side remark, a relation is both LfQuasiRefl and C-LfQuasiRefl iff it is “right-constant”, i.e. it satisfies ∀ x, y , y ∈ X.xRy ↔ xRy . Such a relation also satisfies 2-and 6-AntiTrans, 1-, 3-, 7-, 8-, C- E-Dense, 3-, C-LfEucl, 1-, 8-LfQuasiRefl, 1-, 2-, 3-,7-, C-SemiOrd1, 2-, 3-, C-SemiOrd2, and 1-, 2-, 3-, 8-, 9-, C-, D-Trans. Dually, R is5-LfQuasiRefl and A-LfQuasiRefl iff it is left-constant (defined similarly). A relation isLfQuasiRefl and A-LfQuasiRefl iff it is totally constant, i.e. it satisfies ∀ x , x , y , y ∈ X. x Ry ↔ x Ry .It seems promising to investigate more classes of relations that are characterizable byconjunctions of lifted properties. 55 emma 49. (3-Implications group 1)
1. ( ) Sym ∧ SemiOrd1 → ) Trans ∧ Sym → ) Sym ∧ LfSerial → ) Sym ∧ LfUnique → cnyy ) ASym ∧ LfSerial → qudq ) LfSerial ∧ Trans → uivk ) On a set of (cid:62) ∧ → uklu ) On a set of (cid:62) ∧ → Proof.
1. If R is symmetric, then it coincides with its symmetric kernel R ; hence,if the former is SemiOrd1, so is the latter.2. Similar to 1.3. Similar to 1.4. Similar to 1.5. Given y , we find some x with xRy by LfSerial, this implies ¬ yRx by ASym, hence xR y .6. Let xR y hold; w.l.o.g. let xRy hold. By left seriality of R , we find some x (cid:48) suchthat x (cid:48) Rx ; by transitivity, this implies x (cid:48) Ry . Therefore, xR x (cid:48) and x (cid:48) R y .7. Assume for contradiction xR y and yR z , but xR z . That is, we have exactlyone of xRy and yRx , and similar for y, z and for x, z . Of the 8 cases, only 2satisfy LfUnique, viz. those corresponding to directed cycles. By 8-Trans, any fourthelement w must be comparable to (w.l.o.g.) x and y ; this is impossible due toLfUnique.8. Assume for contradiction xR y and yR z , let w (cid:54) = w (cid:48) be both distinct from x, y, z .From 3-SemiOrd2, we obtain that w is related (w.r.t. R ) to one of x, y, z . By3-LfUnique, neither wRz nor wRy can hold, so the only case that could violate 2-SemiOrd2 is xRw ∧ wRx . By the same argument, we obtain xRw (cid:48) ∧ w (cid:48) Rx . However, wRx ∧ w (cid:48) Rx violates 3-LfUnique. (cid:3) Lemma 50. (3-Implications group 2)
1. ( ) Sym ∧ → AntiTrans2. ( ) Sym ∧ → Dense3. ( ) Sym ∧ → LfSerial4. ( ) Sym ∧ → LfUnique5. ( ) Sym ∧ → LfQuasiRefl6. ( xzog ) LfQuasiRefl ∧ → Proof.
1. If R is symmetric, then its symmetric kernel R coincides with R itself;hence, if the former is anti-transitive, so is the latter.2. Similar to 1, using the symmetric closure instead of the kernel.3. Similar to 1.4. Similar to 1.5. Similar to 1. 56. Let x, y and y, z be incomparable w.r.t. R . Then x, x and y, y are incomparablew.r.t. R by 8-LfQuasiRefl. Assume for contradiction xRz ∨ zRx . In the first case,we have the contradiction xRx by 3-LfQuasiRefl. In the second case, we have zRz by 3-LfQuasiRefl; but y, y and y, z incomparable should imply z, z incomparable. (cid:3) Lemma 51. (lu, de)
Let
Sym = { , , , , , , E, F } denote the set of all unary opera-tions that are guaranteed to yield always a symmetric relation by Lem. 10.14. Let p, q, r be unary operations such that1. ( p ∧ q ) = 0,2. ( p ∨ q ) ⊇ ( ¬ r ), and3. ( ¬ r ) ∈ Sym , or p, q ∈ Sym .Then p -LfUnique and q -LfUnique implies r -Dense. Proof.
First, we prove the case ( ¬ r ) ∈ Sym . Assume for contradiction that xR r z , but(using symmetry of R ¬ r ) that yR ¬ r x ∨ yR ¬ r z for all y . Let p (cid:48) = ( p ∧ ¬ r ) and q (cid:48) = ( q ∧ ¬ r ),then p (cid:48) and q (cid:48) are disjoint like p and q , and ( ¬ r ) = ( p (cid:48) ∨ q (cid:48) ), so we can make the followingcase distinction for all y :1. If yR p (cid:48) x , then yR p x by Lem. 7.4;2. if yR q (cid:48) x , then similarly yR q x ;3. if yR p (cid:48) z , then yR p z ;4. if yR q (cid:48) z , then yR q z .Choosing 5 distinct elements y , . . . , y , one of the cases must appear twice. Doubleappearance of case 1 or 3 contradicts p -LfUnique, double appearance of case 2 or 4 con-tradicts q -LfUnique.For the case p, q ∈ Sym , we use a similar reasoning: Assume for contradiction that xR r z , but xR ¬ r y ∨ yR ¬ r z for all y . Defining p (cid:48) , q (cid:48) as above, we get these cases:1. If xR p (cid:48) y , then xR p y by Lem. 7.4, hence yR p x by symmetry of R p ;2. if yR q (cid:48) x , then similarly yR q x ;3. if yR p (cid:48) z , then yR p z ;4. if yR q (cid:48) z , then yR q z .Again, double appearance of 1 or 3 contradicts p -LfUnique, and double appearance ofcase 2 or 4 contradicts q -LfUnique. (cid:3) Lemma 52. (3-Implications group 5)
1. ( gtlu ) C-SemiOrd2 ∧ SemiOrd2 → uydr ) On a domain of (cid:62) ∧ → Proof.
1. The proof extensively uses boolean connectives on operations (Lem. 7) toshorten the proof presentation. Let x R x and x R x , and let w be given; we haveto show wR x i for some i ∈ { , , } . Assume for contradiction there is no such i ,that is, wR x i for all i .. We distinguish two cases:57 If wRw ,then wRw and wRw implies wR y for each y , by SemiOrd2. Hence wR x i foreach i by assumption. But x R C x and x R C x requires wR E x i for some i , byC-SemiOrd2, which is a contradiction. • If ¬ wRw ,then wR C w and wR C w implies wR E y for each y , by C-SemiOrd2. Hence, wR x i for each i by assumption. But this contradicts SemiOrd2.2. Follows from Lem. 51 with p = 3, q = 4, and r = 8, since ( ¬ r ) = 7 ∈ Sym . (cid:3) Lemma 53. (3-Implications group 6)
1. ( vxfj ) 7-LfUnique ∧ → ASym2. ( xlfr ) 1-LfQuasiRefl ∧ C-Trans → SemiOrd1
Proof.
1. Let xRy , assume for contradiction yRx , i.e. xR y . By 6-LfSerial, we obtainsome w such that wR x . Since xR y implies xR y , and similarly wR x implies xR w , we get w = y by 7-LfUnique. But xR y and xR y (using symmetry of R )implies xR y by Lem. 7.2, a contradiction.2. Let wRx and xR y and yRz . Assume for contradiction ¬ wRz . Then also zRy ,since else, we had ¬ wRy by C-Trans, and ¬ wRx by C-Trans again. Applying 1-LfQuasiRefl, we obtain yR y . But ¬ xRy and ¬ yRx imply the contradiction ¬ yRy by C-Trans. (cid:3) Lemma 54. (3-Implications group 7)
1. ( ewvq ) 2-SemiOrd1 ∧ → g0rw ) 2-SemiOrd2 ∧ → C-SemiOrd23. ( voye ) On a domain of (cid:62) ∧ → C-Dense4. ( voyr ) On a domain of (cid:62) ∧ → wfyp ) On a domain of (cid:62) ∧ → Proof.
1. Let xR y , w.l.o.g. let xRy . By 2-LfSerial, we obtain x (cid:48) , y (cid:48) with x (cid:48) R x and y (cid:48) R y . We have two cases: • If x, y (cid:48) are comparable w.r.t. R , then xR y (cid:48) , that is, y (cid:48) is an intermediateelement, and we are done. • Else, we apply 2-SemiOrd1 to x (cid:48) R x , x, y (cid:48) incomparable w.r.t. R , y (cid:48) R y toobtain x (cid:48) R y , hence x (cid:48) R y , hence x (cid:48) being an intermediate element.2. Assume for contradiction ¬ xRy ∧ ¬ yRz , but wRx ∧ xRw ∧ wRy ∧ yRw ∧ wRz ∧ zRw for some x, y, z, w . By 7-Trans, we get xRy ∨ yRx and yRz ∨ zRy , that is, yRx and zRy . That is, zR y ∧ yR x , but w is incomparable (w.r.t. R ) to x , y , and 2,contradicting 2-SemiOrd2.3. Follows from Lem. 51 with p = 1, q = 6, and r = C , since p, q ∈ Sym .4. Follows from Lem. 51 with p = 1, q = 6, and r = 8. since ( ¬ r ) = 7 ∈ Sym .5. Follows from Lem. 51 with p = 1, q = 8, and r = 6, since ( ¬ r ) = 9 ∈ Sym . (cid:3) emma 55. (cjdj) If R is ASym, then R is CoRefl, LfEucl, LfUnique, Sym, AntiTrans,ASym, AntiSym, Trans, SemiOrd1, Dense ( cjdj ). Proof.
From the assumption follows that R is empty, hence has all claimed propertiesby [1, Exm. 74, p.48]. (cid:3) Theorem 56. ( -implication axioms) Recall that a “3-implication” is a formula of theform ∀ R. lprop ( R ) ∧ lprop ( R ) → lprop ( R ), where lprop i are unnegated lifted properties.Consider the inference system from Def. 25 and 26; inferences are understood to applythe equivalences from Thm. 22 as needed.A 3-implication is valid (w.r.t. a relation domain of (cid:62) Proof. “ ⇐ ”: All axioms have been proven to be valid, by EProver, or manually(Sect. 5.8), or both. The inference rules of Def. 25 and 26 are obviously sound.“ ⇒ ”: All 3-implications that couldn’t be inferred were proven to be invalid, either bya computer-generated counter-example on a domain of 5 elements, or a manual counter-example w.r.t. a finite (Sect. 5.4) or infinite (Sect. 5.5) domain. (cid:3) . References [1] J. Burghardt, Simple Laws about Nonprominent Properties of Binary Relations, Tech-nical Report, URL https://arxiv.org/abs/1806.05036v2 , 2018.[2] A. K. Sen, Quasi-Transitivity, Rational Choice and Collective Decisions, Review ofEconomic Studies 36 (3) (1969) 381–393, URL