Hilbert Calculus. Two kinds of calculi:
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1 Hilbert Calculus Two kinds of calculi: Calculi as basis for automatic techniques Examples: Resolution, DPLL, BDDs Calculi formalizing mathematical reasoning (axiom, hypothesis, lemma..., derivation ) Examples: Hilbert Calculus, Natural Deduction 1
2 Resolution Calculus vs. Hilbert Calculus Resolution calculus Hilbert calculus Proves unsatisfiability Proves consequence (F 1,...,F n = G) Formulas in CNF Formulas with und Syntactic derivation Syntactic derivation of F 1,...,F n G of the empty clause from F from axioms and hypothesis Goal: Goal: automatic proofs model mathematical reasoning Completeness proof Completeness proof comparatively simple comparatively involved 2
3 Recall: Consequence A formula G is a consequence or follows from the formulas F 1,...,F k if every model A of F 1,...,F k that is suitable for G is also a model of G If G is a consequence of F 1,...,F k then we write F 1,...,F k = G. 3
4 Preliminaries In the following slides, formulas contain only the operators und. Recall: F G F G und F G (F G). The calculus defines a syntactic consequence relation (notation: F 1,...,F n G), intended to mirror semantic consequence. We will have: F 1,...,F n G iff F 1,...,F n = G (syntactic consequence and semantic consequence will coincide). 4
5 Axiom schemes We take five axiom schemes or axioms, with F,G as place-holders for formulas: (1) F (G F) (2) (F (G H)) ((F G) (F H)) (3) ( F G) (G F) (4) F ( F G) (5) ( F F) F An instance of an axiom is the result of substituting the place-holders of the axiom by formulas. Easy to see: all instances are valid formulas. Example: Instance of (4) with A B and C for F and G: ( A B) ( ( A B) C) 5
6 Derivations in Hilbert calculus Let S be a set of formulas - also called hypothesis - and let F be a formula. We write S F and say that F is a syntactic consequence of S in Hilbert Calculus if one of these conditions holds: Axiom: Hypothesis: Modus Ponens: F is an instance of an axiom F S S G F and S G, i.e. both G F and G are syntactic consequences of S. 6
7 Modus Ponens Derivation rule of the calculus, allowing to generate new syntactic consequences from old ones: S S S G F G F 7
8 Example of derivation 1. A ((B A) A) Instance of Axiom (1) 2. (A ((B A) A)) ((A (B A)) (A A)) Instance of Axiom (2) 3. (A (B A)) (A A) Modus Ponens with 1. & A (B A) Instance of Axiom (1) 5. A A Modus Ponens with 3. & 4. Remark: The same derivation works for arbitrary formulas F,G instead of A,B, and so we can derive F F for any formula F. We can therefore see a derivation as a way of producing new axioms (the axiom F F in this case). 8
9 Correctness and completeness Correctness: If F is a syntactic consequence from S, then F is a consequence of S. Completeness: If F is a consequence of S, then F is a syntactic consequence from S. 9
10 Correctness proof of the Hilbert calculus Correctness Theorem: Let F be an arbitrary formula, and let S be a set of formulas such that S F. Then S = F. Proof: Easy induction on the length of the derivation of S F. 10
11 Completeness proof: preliminaries Wie wish to prove: if S = F, then S F. How could this work? Induction on the derivation? there is no derivation! Induction on the structure of the formula F? For the induction basis we would have to prove for an atomic formula A: if S = A then S A. But how do we construct a derivation of S A if all we know is S = A? 11
12 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) 12
13 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) (2) Definition: S is inconsistent if there is a formula F such that S F and S F. 12
14 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) (2) Definition: S is inconsistent if there is a formula F such that S F and S F. (3) S F iff S { F } is inconsistent. (To be proved!) 12
15 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) (2) Definition: S is inconsistent if there is a formula F such that S F and S F. (3) S F iff S { F } is inconsistent. (To be proved!) (4) Unsatisfiable sets are inconsistent. (To be proved!) 12
16 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) (2) Definition: S is inconsistent if there is a formula F such that S F and S F. (3) S F iff S { F } is inconsistent. (To be proved!) (4) Unsatisfiable sets are inconsistent. (To be proved!) Proof sketch: Assume S = F. Then S { F } is unsatisfiable by (1). Then S { F } is inconsistent by (4). Then S F by (3). 12
17 Completeness - Proof sketch (1) S = F iff S { F } is unsatisfiable. (Trivial) (2) Definition: S is inconsistent if there is a formula F such that S F and S F. (3) S F iff S { F } is inconsistent. (To be proved!) (4) Unsatisfiable sets are inconsistent. (To be proved!) Proof sketch: Assume S = F. Then S { F } is unsatisfiable by (1). Then S { F } is inconsistent by (4). Then S F by (3). We prove (3) und (4). 12
18 (In)consistency Definition: A set S of formulas is inconsistent if there is a formula F such that S F and S F, otherwise it is consistent. Observe: inconsistency is a purely syntactic notion!! 13
19 Examples of inconsistent sets {A, A} { (A (B A))} { B, B B} {C, ( C D)} 14
20 Important tool: the Deduction Theorem Theorem: S {F } G iff S F G. Proof: Assume S F G. Then S {F } F G. Using S {F } F and Modus Ponens we get S {F } G. Assume S {F } G. Proof by induction on the derivation (length): Axiom/Hypothesis: G is instance of an axiom or G S {F }. If F = G use example of derivation to prove S F F. Otherwise S G and by Axiom (1) S G (F G). By Modus Ponens we get S F G. Modus Ponens: Then S {F } G is derived by Modus Ponens from some S {F } H G and S {F } H. By ind. hyp we have S F (H G) and S F H. From Axiom (2) we get S (F (H G)) ((F H) (F G)). Modus Ponens yields S F G. 15
21 Consequences of the Deduction Theorem Lemma I: S { F } F iff S F Proof: Assume S { F } F holds. By the Deduction Theorem S F F. Using Axiom (5) we get S ( F F) F. By Modus Ponens we get S F. The other direction is trivial. 16
22 Completeness - Proof of (3) Assertion (3): S F iff S { F } is inconsistent. Proof: Assume S F. Then S { F } F. Since S { F } F, the set S { F } is inconsistent. Assume S { F } is inconsistent. Then there is a formula G s.t. S { F } G and S { F } G. By Axiom (4) we get S { F } G ( G F). Two applications of Modus Ponens yield S { F } F. Lemma I yields S F. 17
23 Completeness - Proof of (4) Recall assertion (4): Unsatisfiable sets are inconsistent. We prove the equivalent assertion: Consistent sets are satisfiable. How do we prove an assertion like this? 18
24 Completeness - Proof of (4) Recall assertion (4): Unsatisfiable sets are inconsistent. We prove the equivalent assertion: Consistent sets are satisfiable. How do we prove an assertion like this? Answer: Construct a satisfying truth assignment A as follows: If A S then set A(A) := 1. If A S then set A(A) := 0. 18
25 Completeness - Proof of (4) Recall assertion (4): Unsatisfiable sets are inconsistent. We prove the equivalent assertion: Consistent sets are satisfiable. How do we prove an assertion like this? Answer: Construct a satisfying truth assignment A as follows: If A S then set A(A) := 1. If A S then set A(A) := 0. Problem: What do we do if neither A S nor A S? 18
26 Perhaps we can avoid the problem? Definition: A set S of formulas is maximally consistent if it is consistent and for every formula F either F S or F S. 19
27 Perhaps we can avoid the problem? Definition: A set S of formulas is maximally consistent if it is consistent and for every formula F either F S or F S. We extend S to a maximally consistent set S S. 19
28 Completeness - Proof sketch for (4) (4) Consistent sets are satisfiable. 20
29 Completeness - Proof sketch for (4) (4) Consistent sets are satisfiable. (4.1) Every consistent set can be extended to a maximally consistent set. 20
30 Completeness - Proof sketch for (4) (4) Consistent sets are satisfiable. (4.1) Every consistent set can be extended to a maximally consistent set. (4.2) Let S be maximally consistent and let A be the assignment given by A(A) = 1 if A S and A(A) = 0 if A / S. Then A satisfies S. 20
31 Proof of (4.1) - Preliminaries Lemma II: Let S be a consistent set and let F be an arbitrary formula. Then: S {F } or S { F } (or both) are consistent. Proof: Assume S is consistent but both S {F } and S { F } are inconsistent. Since S { F } is inconsistent we have S F by Assertion (3). Since S {F } is inconsistent there is a formula G s.t. S {F } G and S {F } G, and the Deduction Theorem yields S F G and S F G. Modus Ponens yields S G and S G. This contradicts the assumption that S is consistent. 21
32 Proof of (4.1) Assertion (4.1): Every consistent set can be extended to a maximally consistent set. Proof: Let F 0,F 1,F 2... be an enumeration of all formulas. Let S 0 = S and S i+1 = { Si {F i } if S i {F i } consistent S i { F i } if S i { F i } consistent (this is well defined by Lemma II) By definition, every S i is consistent. Let S = i=1 S i. If S were inconsistent, some finite subset would also be inconsistent. So S is consistent. By definition, S is maximally consistent. 22
33 Proof of (4.2) - Preliminaries Lemma III: Let S be a maximally consistent set: (1) For every formula F: F S iff S F. (2) For every formula F: F S iff F S. (3) For every two formulas F,G: F G S iff F S or G S. Proof: We prove only: if F S then F G S (others similar). From F S we get: 1. S F because F S 2. S F ( G F) Axiom (1) 3. S G F Modus Ponens to 1. & S ( G F) (F G) Axiom (3) 5. S F G Modus Ponens to 3. & 4. 23
34 Proof of (4.2) Assertion (4.2): Let S by maximally consistent, and let A be the assignment given by: A(A) = 1 iff A S. Then A satisfies S. Proof: Let F be a formula. We prove: A(F) = 1 iff F S. By induction on the structure of F (and using Lemma III): Atomic formulas: F = A. Easy. Negation: F = G. We have: A(F) = 1 iff A(G) = 0 iff G S iff G S iff F S. Implication: F = F 1 F 2. We have: A(F) = 1 iff A(F 1 F 2 ) = 1 iff (A(F 1 ) = 0 or A(F 2 ) = 1) iff (F 1 S or F 2 S) iff F 1 F 2 S iff F S. 24
35 A Hilbert Calculus for predicate logic We extend formulas by allowing universal quantification. Three new axiom schemes: (6) ( x F) F[x/t] for every term t. (7) ( x (F G)) ( x F x G). (8) F x F if x does not occur free in F. Theorem: The extension of the Hilbert Calculus is correct and complete for predicate logic. 25
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