Some important properties Lectures of Prof. Doron Peled, Bar Ilan University

We repeat some notation and definitions. An assignment z satisfies a formula if M PL (, z)=1 |== means that logically implies, i.e., each assignment that satisfies all the formulas in also satisfies. |-- means that we can prove from, i.e., there is a proof sequence that uses assumptions from, the axioms A1, A2, A3 and the proof rule MP, and ends with.

Substitution of variables by formulas Let be a WFF. A function s:Var WFF is called a substitution function. We define inductively a function subst(,s) that replaces in all the variables p i by s(p i ). Basis: subst(p i,s)=s(p i ), subst(F,s)=F, subst(T,s)=T. Closure: subst(( 1 /\ 2 ),s)=(subst( 1,s)/\subst( 2,s)) subst(( 1 \/ 2 ),s)=(subst( 1,s) subst( 2,s)) Etc.

For example s={, } subst((p 1 \/p 2 ),s)=(subst(p 1,s)\/subst(p 2,s))= ((p 3 \/p 4 )\/(p 5 \/p 6 )) Finite dependency of substitution: Let be a WFF with all its propositions inside {p 1,p 2,…,p n }. Let s 1 and s 2 be two substitution functions such that for each i, s 1 (p i )=s 2 (p i ). Then subst(,s 1 )=subst(,s 2 ).

Some propositions If is not satisfiable and s is a substitution, then subst(,s) is not satisfiable. If is a tautology and s is a substitution, then subst(,s) is a tautology.

Provability and satisifiability Lemma(*): |== iff { F} is unsatisfiable. One direction: from |==, any assignment that satisfies must also satisfy, thus cannot satisfy { F}. Conversely, if { F} is unsatisfiable, any assignment that satisfies cannot satisfy { F}, hence it must satisfy.

Consistency A set of formulas is consistent if |-/-F. A set is maximally consistent if it is consistent and for every formula, either or ( F) belongs to the set.

Consistency and satisfiability inconsistent unsatisfiable (=soundness!) Can be proved from soundness: if we can prove F from then since our proof system is sound, F logically follows from. Thus there cannot be a satisfying assignment to our assumptions. This implies consistency, as will be shown later. consistent satisfiable (=completeness!) The contraposition is implied by completeness: If is unsatisfiable then |==F. By the completeness theorem, |--F, i.e., is inconsistent. This implies completeness as will be shown later.

Landscape Consistent Satisfiable Soundness Completeness

Proving that consistent satisfiable Same as completeness theorem: start with and extend into * by enumerating the formulas in some order and adding a formula if the set remains consistent. Then * behaves as a truth assignment and thus gives a satisfying assignment to.

consistent satisfiable implies completeness! We showed that every consistent set is satisfiable. Lemma: If |-/- then { F} is consistent Proof: We show the contraposition: If { F} is inconsistent, then |--. If so, then { F}|-- (since from F we can prove anything). But we can also prove from { }. Thus, by proof by cases, we have |--. Now assume |== but |-/-. Then according to the Lemma above { F} is consistent. But according to lemma (*), { F} is not satisfiable. Contradiction.

Compactness theorem Let be an infinite set of propositional formulas. Then is satisfiable iff every finite subset of is satisfiable. Proof: as shown before, is satisfiable iff is consistent. If is consistent, then every finite subset of it is consistent. Conversely, if is inconsistent, there is a finite subset of it that proves F, hence is inconsistent.

inconsistent unsatisfiable implies consistency Lemma: If |-- then { F} is inconsistent. Proof: If |-- it follows that from { F} one can prove both and { F}, hence one can prove F, which makes { F} inconsistent. Now suppose that |--. Then { F} is inconsistent. Thus, { F} is unsatisfiable. Therefore from Lemma (*) |==.