bv-∃

Theorem.

Arguments:

y (sv), phi (pr), x (sv),

Hypotheses:

(x bound in phi)

Assertions:

(x bound in existsyphi)

Proof:

Hyp Ref Line Expr
Hypo1(x bound in phi)
1bv-¬2(x bound in lnotphi)
2bv-∀3(x bound in forallylnotphi)
3bv-¬4(x bound in lnotforallylnotphi)
df-ex5(existsyphileftrightarrowlnotforallylnotphi)
4, 5bvdef6(x bound in existsyphi)