Some important properties

Lectures of Prof. Doron Peled, Bar Ilan University

We repeat some notation and definitions.

An assignment z satisfies a formula if MPL(, 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:VarWFF is called a substitution function.

We define inductively a function subst(,s) that replaces in all the variables pi by s(pi).

Basis: subst(pi,s)=s(pi), 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={<p1,(p3\/p4)>,<p2,(p5\/p6)>}

subst((p1\/p2),s)=(subst(p1,s)\/subst(p2,s))=

)) p3\/p4)/\(p5\/p6((

Finite dependency of substitution: Let be a WFF with all its propositions inside {p1,p2,…,pn}. Let s1 and s2 be two substitution functions such that for each i, s1(pi)=s2(pi). Then subst(,s1)=subst(,s2).

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 consistentProof: 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 (*) |==.