Monadic Predicate Logic is Decidable. Boolos et al, Computability and Logic (textbook, 4 th Ed.)

Transcription

1 Monadic Predicate Logic is Decidable Boolos et al, Computability and Logic (textbook, 4 th Ed.)

2 These slides use A instead of E instead of & instead of - instead of Nota>on Equality statements are atomic formulas: x=y, a=b, x=a, etc sentence = formula with no free variables

3 Monadic First- Order Predicate Logic (FOPL) The fragment of Predicate logic that uses no predicates with more than 1 argument In: Ex F(x) & Ey - F(y) AxEy - (x=y) (equality statements permised!) G(a) v G(b), etc. Out: AxEy (R(x,y)) R(a,b,b) v Ex F(x) (because they use a dyadic/triadic/etc predicates)

4 Some reasoning tasks For given sentences ϕ and ψ, does ψ follow from ϕ? ( Does ϕ have ψ as a logical consequence? ) More precisely: Is it true that for all models M, if M = ϕ then M = ψ? For given a sentence ϕ, is ϕ sa>sfiable? We mean: Is there a model that M such that M = ϕ? E.g., ExF(x) & Ex- F(x) is sa>sfiable Analogous for formulas with free variables

5 Monadic FOPL sa>sfiability is decidable

6 Key theorem (Lőwenheim- Skolem 1915) If S is a monadic sentence that has a model, then S is true in some model whose domain consist of at most 2 k.r members, where k is the number of predicate lesers in S and r the number of variables in S. Part 1, proof: The Key Theorem Part 2, proof: It follows that monadic FOPL is decidable

7 Proof of Part 1 (= Key Theorem) Let S be a sentence of monadic FOPL. Its predicates are P 1,..,P k Let M = S, and let D be the domain of M D may be infinite Let the signature of d in D (henceforth sig(d)) be the sequence <j 1,..,j k > where j i =1 if M specifies that P i is true of d and j i =0 otherwise sig(d) tell us which predicates in S are true of d given S, there are exactly 2 k different possible signatures

8 We call d and d similar if sig(d)=sig(d ) This means that d and d happen to share all their proper>es P 1,..,P k. similar is an equivalence rela>on, so each d in D belongs to an equivalence class of similar domain elements Each equivalence class is a subset of D there are at most 2 k equivalence classes

9 Towards a smaller model M Construct a subset E D as follows: Choose r elements from each equivalence class If a class has fewer than r elements then choose them all E cannot have more than 2 k.r elements (r for each equivalence class) Define: M is the restric>on of M to E just like M, but defined for elements of E only To be proven: M = S

10 A useful concept: match Informally: Two sequences of elements of E that are of the same length match if their elements are similar and differences within each sequence are respected in the other: Formally: c 1,..,c n matches d 1,..,d n iff 1. c i is similar to d i (for every 1<= i <= n) 2. c i = c j iff d i = d j (for every 1 <= i,j <= n)

11 Example These 2 sequences of domain objects do not match: c 1,..,c n = a,b,a d 1,..,d n = a,b,c If a and c are similar then clause 1 is fulfilled, but clause 2 is not (because c 1 =c 3 but d 1 <>d 3 ) Reason behind clause 2: equality statements in FOPL (e.g. in the sentence AxEy - (x=y))

12 Another useful concept A formula ϕ containing at most the free variables x 1,..,x n is sa?sfied by elements d 1,..,d n in a model M iff M = ϕ (x 1 :=d 1,..,x n :=d n ) (A simple extension of the idea of sa>sfiability)

13 Lemma Let G(x 1,..,x n ) be any subformula of S, containing at most the free variables x 1,..,x n d 1,..,d n a sequence of elements of D (the original domain) e 1,..,e n a sequence of elements of E (dom constructed above) d 1,..,d n matches sequence e 1,..,e n Then G(x 1,..,x n ) is sa>sfied by d 1,..,d n in M iff G(x 1,..,x n ) is sa>sfied by e 1,..,e n in M

14 Why does this lemma hold? (informally) As far as the predicates P 1,..,P k occurring in S are concerned, each element d i is just like e i Clause 1 of match The only other thing that can maser (because of equality statements!) is whether two elements in a given sequence are iden>cal Clause 2 of match

15 Sketch of a formal proof (by formula induc>on) Base Cases: G is atomic. G is of the form P i (t) or of the form t 1 =t 2 (t, t 1, and t 2 are variables or constants) 1. Let G = P i (t). We need to prove: P i (t) is sa>sfied by d 1 in M iff P i (t) is sa>sfied by e 1 in M But d 1 and e 1 are similar, hence the same predicates hold true of d 1 and e 1 (including the predicate P i ). This proves the first Base Case.

16 Sketch of proof by formula induc>on Base Cases: G is atomic. G is of the form P i (t) or of the form t 1 =t Let G = t 1 =t 2. We need to prove t 1 =t 2 is sa>sfied by d 1, d 2 in M iff t 1 =t 2 is sa>sfied by e 1, e 2 in M But the sequences d 1 d 2 and e 1 e 2 match, hence d 1 = d 2 iff e 1 =e 2. This proves the second Base Case.

17 Sketch of proof by formula induc>on Induc?ves Cases: [Proofs omised, but see Ques>ons for the Prac>cal] It suffices to address -, v, A. (1) Assume the Lemma holds for ϕ. Prove that it holds for φ. (2) Assume the Lemma holds for φ and ψ. Prove that it holds for φ v ψ. (3) Assume the Lemma holds for φ. Prove that it holds for Axφ.

18 S is itself a subformula of S, hence it follows directly (with n=0) from the Lemma that S is true in M iff S is true in M Recall: M may be infinite, but M is finite, with at most 2 k.r elements

19 Proof of Part 2 Let S be a FOPL sentence Associate with S a quan>fier- free formula S such that S is sa?sfiable iff S is. (Next page) If we manage to do this then we deduce: The sa>sfiability of S can be decided using truth tables (since these suffice for deciding the sa>sfiability of S ) Hence the sa>sfiability of S can be decided

20 Proof of Part 2 Making use of Part 1, associate with S a quan>fier- free formula S which is sa>sfiable iff S is. As follows: Induc>vely associate a quan>fier- free H with each subformula H of S, as follows: If H is atomic: H =H (no change!) If H is a truth func>onal compound: H =H If H=ExF: H =F(a 1 )v..vf(a m ) m=2 k.r If H=AxF: H =F(a 1 )&..&F(a m ) m=2 k.r S itself is a subformula of S, so this constructs a quan>fier- free S as well. The construc>on guarantees: S is sa>sfiable iff S is sa>sfiable

21 Example (using an arbitrary S) Consider S = ((ExF(x) & ExG(x)) & - (Ex(F(x)&G(x)))) Here k=2, r=3, so 2 k.r=12 The following formula is constructed: F(a 1 ) v.. v F(a 12 ) & G(a 1 ) v.. v G(a 12 ) & - ((F(a 1 )&G(a 1 )) v..v (F(a 12 )&G(a 12 )))

22 Example F(a 1 ) v.. v F(a 12 ) & G(a 1 ) v.. v G(a 12 ) & - ((F(a 1 )&G(a 1 )) v..v (F(a 12 )&G(a 12 ))) Proposi>onal formula with 24 atoms Each can be True or False => truth table has 2 24 rows. Try to find a row that is True. Example: F(a 1 ),F(a 2 ),..F(a 12 ), G(a 1 ),G(a 2 ),G(a 3 ),..G(a 12 ) T F F F T F F

23 Example F(a 1 ),F(a 2 ),..F(a 12 ), G(a 1 ),G(a 2 ),G(a 3 ),..G(a 12 ) T F F F T F F We can read off from this a model with 12 elements that sa>sfies the formula. The same model must sa>sfy the original (quan>fied) formula S too.

24 Concluding The proof suggests an algorithm for deciding whether a formula is sa>sfiable Not sa>sfiable è no row of the truth table is True Also applicable to logical consequence Implementa>ons exist 2 k implies Exponen>al in complexity (though faster methods exist) Decidability proofs oqen tell us something about the worst- case run?me of a program

25 Other FOPL fragments For every n, it is decidable whether a given formula of FOPL has a model of size m <= n Not proven here However, dyadic FOPL is undecidable If >me permits, we will prove this later For now, just one observa>on

26 Observe: Key Theorem does not hold for dyadic FOPL Example: the following FOPL sentence does not have a finite model Ex(x=x) & AxEy(x<y) & Axyz((x<y & y<z) à x<z) & Ax- (x<x) Why not?