# Techniques for Proving the Completeness of a Proof System Hongseok Yang Seoul National University Cristiano Calcagno Imperial College.

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Techniques for Proving the Completeness of a Proof System Hongseok Yang Seoul National University Cristiano Calcagno Imperial College

Completeness Question Completeness: Is a proof system powerful enough to verify all true facts? Proof system for classical propositional logic: ` P ) (Q ) R) ` (P ) (Q ) Q’)) ) ((P ) Q) ) (P ) Q’)) ` :: P ) P ` P ` P ) Q ` Q Truth: P holds (denoted ² P) iff P always evaluates to true by the “table method.” Completeness Theorem: if ² P, then ` P. Exercise: Prove ` ((q ) r) ) q) ) q.

Reasons for Studying Completeness 1. Sometimes it is easier to show the truth of a formula than to derive the formula. 2. The completeness result shows that nothing is missing in a proof system. 3. The completeness result formalizes what a proof system achieves. 4. With a completeness result, a paper about a proof system has more chances to get accepted.

Goal of My Talk To present common techniques for showing the completeness, so that you can apply them to your own problem. In particular, to explain the following concepts: maximally consistent set truth lemma Lindenbaum lemma If time permits, I will briefly explain what I’m working on with Calcagno in Imperial college.

Simple Modal Logic P := q | : P | P ) P |  P Proof system: usual rules in classical logic with the following additional ones for the modality: ` P `  P `  (P ) Q) ) (  P )  Q) Example: student, ð phd, ðð professor deadlock, ð deadlock, : ð : deadlock

Semantics A model M is a triple (M, R:M \$ M, I:Symb !P (M)). Interpretation of Simple Modal Logic M,m ² q iff m 2 I(q) M,m ² : P iff M,m 2 P M,m ² P ) Q iff if M,m ² P, then M,m ² Q M,m ²  P iff for all n, if R(m,n), then M,n ² P Example: M=years, R(n,m) iff m=n+1, I(phd)={2001,…}, I(student)={1982,…,2001}

Completeness Question P is valid (denoted ² P) iff for all models (M,R,I) and all m in M, (M,R,I),m ² P. Completeness: If ² P, then ` P. General guide: consider the contrapositive! Contrapositive: if 0 P, then 2 P. Guide: for each P such that 0 P, construct a model (M,R,I) with m in M such that (M,R,I),m ² : P.

Strategy for Constructing a Required Model Build a model M = (M,R,I) such that 1. [term model] each m in M is a set of formulas; 2. [truth lemma] for all m and Q, m ² Q iff Q is in m; 3. there exists n in M containing : P. This model is what we are looking for. (Why?) How to build such a model?

Inferring Requirements from the Truth Lemma Let M =(M,R,I) be a model such that each m in M is a set of formulas. Try to use induction to show the truth lemma for M : for all m in M and Q, m ² Q iff Q is in m. What conditions do R and I satisfy?

Inferring Requirements from the Truth Lemma Let M =(M,R,I) be a model such that each m in M is a set of formulas. Try to use induction to show the truth lemma for M : for all m in M and Q, m ² Q iff Q is in m. What conditions do R and I need to satisfy? 1. Q is not in m iff : Q is in m. 2. If both Q and Q ) Q’ are in m, then Q’ is in m. 3. If R(m,n) and ð Q is in m, then Q is in n. 4. m is in I(q) iff q is in m.

Maximally Consistent Set A set m of formulas is maximally consistent iff 1. for all Q, only one of Q and : Q is in m; and 2. if Q 0,Q 1,...,Q n 2 m and ` Q 0 ) Q 1 ) … ) Q n ) Q’, then Q’ in m. By turning the conditions into the definition (almost directly), we can construct the required model: M consists of maximally consistent sets of formulas; R(m,n) iff for all ð Q in m, Q in n; m 2 I(p) iff p 2 m. The proof for the truth lemma “almost” works.

Lindenbaum Lemma Still need to show two facts: If m ²  Q, then  Q is in m. There exists m in M such that m ² : P. Lindenbaum Lemma: Let {Q 0,Q 1,…,Q n } be a set of formulas. If 0 Q 0 ) Q 1 ) … ) Q n ) false, then there is a maximally consistent set m s.t. Q 0, Q 1, …, Q n 2 m. Try to show the two properties with Lindenbaum Lemma.

Summary We constructed “canonical” model (M,R,I): M consists of maximally cons. sets of formulas. R(m,n) iff for all  P in m, P is in n. m 2 I(p) iff p in m. The model satisfies the following properties: truth lemma: m ² P iff P is in m. If 0 : P, then there is m such that P 2 m. These two properties lead to the completeness.

My Work with Calcagno Interested in the completeness of Boolean BI wrt PCM models. Conceptual implication: supports that BBI is a logic for computational resources. “Practical” implication: shows that the proof rules in separation logic are powerful enough. Roughly, the question is similar to asking whether our model logic are complete wrt injective models. (M,R,I) is injective iff R(m,n) Æ R(m,n’) ) n=n’ We found a method to transform the “canonical” model to an injective one, while preserving the satisfiability of formulas.

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