CS 486: Spring 2001 Cornell University Prof. Robert Constable

Feb 20 Lecture Notes: Hintikka

∀S : 𝔹 × F orm → 𝔹. Hintikka(S) ⇒ ∃v : V ar → 𝔹. ∀X : F orm.(S(X) ⇒ svalue(X, v) = t)

Proof: Define v as follows for p ∈ V ar

v(p) = S(< T, p >) i.e. assign t top if T p is in S.

(this assigns f if < F, p >∈ S and if neither < T, p > nor < F, p > is in S).

Claim ∀X : F orm.(S(X) ⇒ svalue(X, v) = t)
Proof: by induction on F orm.

base: X is a signed variable; if < T, p >∈ S then svalue(X, v) = t, since v(p) = t. if < F, p >∈ S then svalue(X, v) = t, since v(p) = f. induction: assume the result for all U, V subformulas of X

Case X is an α formula: then α_{1}, α_{2} are both subformulas of α, and since S is Hintikka, S(α_{1}) and S(α_{2}). So by induction, svalue(α_{1}, v) = t and

svalue(α_{2}, v) = t, thus svalue(α, v) = t.

X is a β formula: then β_{1}, β_{2} are both subformulas of β, and since S is Hintikka, one of them is in S, say S(β_{i}). Thus by induction svalue(β_{i}, v) = t, hence svalue(β, v) = t.

Qed