CS 486: Spring 2001 Cornell University Prof. Robert Constable

Feb 15 Lecture Notes: Consistency

Theorem ∀X : F orm. ∀t : Tableau_{X}. ∀v_{0} : V ar → 𝔹.

(svalue(root(t), v_{0}) = t) ⇒

∃Θ : P ath(t). ∀Y : F orm.(Y on Θ ⇒ svalue(Y, v_{0}) = t)

Proof:  by induction on the depth of t.
base:  t is a leaf, the root, then the root is the path Θ.
induction: suppose there is a path Θ_{0} in any tableau t_{n} of depth n;

consider now a tree of depth n + 1.

If Θ_{0} is a path to a leaf in t_{n+1}, then we are done, take Θ = Θ_{0}.

If Θ_{0} is extended in t_{n+1}, then it is either:

(A) (Θ_{0}, α_{1}) or (Θ_{0}, α_{2}) or (B) (Θ_{0}, β_{1}) or (Θ_{0}, β_{2})

In case (A), α is true since it is on Θ_{0} and so are α_{1}, α_{2} true, thus Θ = (Θ_{0}, α_{i})

are possible paths. In case (B), β is true since it is on Θ_{0} and so one of (Θ_{0}, β_{1}), (Θ_{0}, β_{2}) is a

possible path. Pick the true one as the new path, Θ.

Qed

Lemma  ∀X : F orm. ∀t : Tableau_{X}. closed(t) ⇒ ∀v_{0} : V ar_{X} → 𝔹.

¬ ∃Θ : P ath(t). ∀X : F orm.(Y on Θ ⇒ svalue(Y, v_{0}) = t)

Proof:  If t is closed, then on every path Θ there appears T Y and F Y for some Y.

So not all formulas on Θ can be true, no matter what the interpretation v_{0}.

Qed

Corollary  ∀X : F orm. ∀t : Tableau_{X}. closed(t) ⇒
∀v_{0} : V ar_{X} → 𝔹. (svalue(F X, v_{0}) = f).
Theorem  ∀X : F orm. (∃t : Tableau_{X}.closed(t)) ⇒ tautology(X)