aa r X i v : . [ m a t h . L O ] A ug A constructive proof of a theorem byFerreira-Zantema
Toshiyasu AraiGraduate School of Mathematical SciencesUniversity of Tokyo3-8-1 Komaba, Meguro-ku, Tokyo 153-8914, [email protected]
Abstract
This note was written in Jan. 23, 2015 to answer a problem raisedby G. Moser, who asked a constructive proof of a theorem by Ferreira-Zantema[1].
For a binary relation < on a set T , W ( < ) denotes the well-founded part of T with respect to < .Let F be a non-empty and finite set of function symbols with F = N > f has a fixed arity ar ( f ) ∈ ω . When writing f ( a ), we tacitly assume that a ∈ T ( F ) ar ( f ) . If a = ( a , . . . , a n − ), then t ∈ a iff t = a i for an i < n .Let < be a proper order, i.e., irreflexive and transitive relation on T ( F ), andfor each f ∈ F , a relation (not necessarily a proper order) < f on T ( F ) ar ( f ) isgiven.Assume the following three conditions for f ∈ F :1. < contains the subterm relation: for any proper subterm s of t , s < t .2. If f ( b ) < f ( a ), then either f ( b ) ≤ a i for some a i ∈ a , or b < f a .3. ∀ a ∈ T ( F ) ar ( f ) [ a ⊂ W ( < ) ⇒ a ∈ W ( < f )] (1) Theorem 1 (Ferreira-Zantema[1]) T ( F ) ⊂ W ( < ) . We show the following lemma first.
Lemma 2
Let { f , . . . , f K } ⊂ F be a set of distinct function symbols, and a i ∈ T ( F ) ar ( f i ) . Let for i = 0 , , . . . , K , T f, := T ( F ) ar ( f ) T f,i := { b ∈ T ( F ) ar ( f ) : f ( b ) < f i ( a i ) } ( i = 0) A f,i ( b ) : ⇔ [ b ∈ T f,i → b ⊂ W ( < ) → f ( b ) ∈ W ( < )] IH f,i ( a ) : ⇔ ∀ b < f a A f,i ( b )1 ssume the following three for any i = 1 , . . . , K :1. f i +1 ( a i +1 ) < f i ( a i ) .2. IH f i ,i − ( a i ) .3. a i ⊂ W ( < ) .Then f K ( a K ) ∈ W ( < ) . Proof by (meta)induction on N − K .By MIH we have for any f
6∈ { f , . . . , f K } , ∀ b [ f ( b ) < f K ( a K ) → IH f,K ( b ) → b ⊂ W ( < ) → f ( b ) ∈ W ( < )], i.e., ∀ b [ ∀ c < f b A f,K ( c ) → A f,K ( b )]. In otherwords, the Π -predicate A f,K is progressive with respect to < f . Hence Π -BI ⊢ W ( < f ) ⊂ A f,K .We show that s < f K ( a K ) → s ∈ W ( < )by subsidiary induction on the size of terms s .Let s = f ( b ) < f K ( a K ). For any b i ∈ b , b i < f K ( a K ) since < contains thesubterm relation and transitive. By SIH we have b ⊂ W ( < ).First consider the case f = f i for some 1 ≤ i ≤ K . Then by (1) and thetransitivity of < , we have s = f i ( b ) < f i ( a i ).If s ≤ a ij for an a ij ∈ a i , then (3) yields s ≤ a ij ∈ W ( < ), and s ∈ W ( < ).Otherwise we have b < f a i . IH f i ( a i ), (2), with b ⊂ W ( < ) yields s = f i ( b ) ∈ W ( < ).Second consider the case f
6∈ { f , . . . , f K } . By MIH we have W ( < f ) ⊂ A f,K . b ⊂ W ( < ) with (1) yields b ∈ W ( < f ), and A f,K ( b ). On the other hand we have f ( b ) < f K ( a K ), i.e., b ∈ T f,K . Thus we conclude f ( b ) ∈ W ( < ). ✷ From Lemma 2 we see that for any f ∈ F , ∀ a [ IH f, ( a ) → a ⊂ W ( < ) → f ( a ) ∈ W ( < )]. In other words, the Π -predicate A f, is progressive with respectto < f . Hence (Π -BI ⊢ ) W ( < f ) ⊂ A f, , i.e., a ∈ W ( < f ) → a ⊂ W ( < ) → f ( a ) ∈ W ( < ). Under (1) this is equivalent to ∀ a [ a ⊂ W ( < ) → f ( a ) ∈ W ( < )]Now Theorem 1, t ∈ W ( < ) is seen by induction on the size of terms t . Thus weget the following Theorem 3. Theorem 3
Theorem 1 is proved in Π -BI for each finite F . When F ranges over finite sets of function symbols in Theorem 1, we needto prove Lemma 2 by formal complete induction, and the lemma is a sentenceof the form ∀ n ∃ X ∀ Y θ with a first-order θ . Therefore the proof of the lemmais formalizable in Π -BI with full induction schema, but not in Π -BI withrestricted induction. 2ndeed, Theorem 1 for any finite set F implies the well-foundedness W ( ϑ (Ω ω ))of the proof-theoretic ordinal ϑ (Ω ω ) of Π -BI (Lemma 6 below), and is not prov-able in Π -BI . It is known that Kruskul’s theorem is equivalent to W ( ϑ (Ω ω ))over ACA , cf. [2]. Therefore we obtain the following Theorem 4. Theorem 4
Over
ACA the following facts are equivalent each other:1. Any simplification order over any finite F is terminating.2. Theorem 1 for any finite set F .3. W ( ϑ (Ω ω )) .4. Kruskul’s theorem KT ( ω ) for any finite trees. Theorem 5
Over
ACA the following facts are equivalent each other:1. Any simplification order over each finite F is terminating.2. Theorem 1 for each finite set F .3. W ( ϑ (Ω k )) for k = 0 , , , . . .
4. Kruskul’s theorems KT ( k ) for k -branching trees for k = 0 , , , . . . Lemma 6
Theorem 1 for any finite set F implies the well-foundedness W ( ϑ (Ω ω )) . Proof .Each ordinal < ϑ (Ω k +1 ) is represented by a term over the symbols 0 , , . . . , k, + , ϑ and Ω. Let F k = { f i : i ≤ k } ∪ { g, } , where ar ( f i ) = i + 1, ar ( g ) = 2and ar (1) = 0. Each ground term t ∈ T ( F ) denotes a non-zero ordinal o ( t ) < ϑ (Ω k +1 ) as follows. o (1) = 1, o ( g ( t, s )) = o ( t ) o ( s ) for the naturalsum α β of ordinals α and β , and o ( f i ( t i , . . . , t )) = ϑ (Ω i o ( t i ) + · · · + Ω o ( t )).Let t ≺ s : ⇔ o ( t ) < o ( s ) . The relation ≺ on terms is a proper order, and contains the subterm relation. α = ϑ (Ω i α i + · · · + Ω α ) < ϑ (Ω i β i + · · · + Ω β ) = β iff either α ≤ β j for a j ,or α j < β for any j and ( α i , . . . , α ) < lx ( β i , . . . , β ) for the lexicographic order < lx . Therefore if f i ( t i , . . . , t ) ≺ f i ( s i , . . . , s ), then either f i ( t i , . . . , t ) (cid:22) s j for a j , or ( t i , . . . , t ) ≺ lx ( s i , . . . , s ), where ≺ lx is the lexicographic order ontuples of the same lengths induced by ≺ . Moreover if g ( t , t ) ≺ g ( s , s ), then( t , t ) ≺ m ( s , s ) : ⇔ ∃ i, j < t i ≺ s j ∧ t − i (cid:22) s j − ), where ≺ m is a multisetextension of ≺ . Hence (1) is enjoyed for each function symbol. Theorem 1 yields T ( F k ) ⊂ W ( ≺ ).For each ordinal (term in normal form) α < ϑ (Ω k +1 ), let α + denote theordinal obtained from α by replacing the subterms 0 by 1. 0 + = 1, ( α β ) + = α + β + and ( ϑ (Ω i α i + · · · + Ω α )) + = ϑ (Ω i α + i + · · · + Ω α +0 ). Then it is easyto see that α < β iff α + < β + . Moreover for each α < ϑ (Ω k ) there exists a term t such that o ( t ) = α + . Therefore T ( F k ) ⊂ W ( ≺ ) yields W ( ϑ (Ω k +1 )). ✷ eferenceseferences