sb-spec-ex

Theorem.

Arguments:

a (st), x (sv), phi (pr),

Assertions:

([a/x]phitoexistsxphi)

Proof:

Hyp Ref Line Expr
spec1(forallxlnotphito[a/x]lnotphi)
sb-neg2([a/x]lnotphileftrightarrowlnot[a/x]phi)
1, 2brpi223(forallxlnotphitolnot[a/x]phi)
3con>4([a/x]phitolnotforallxlnotphi)
df-ex5(existsxphileftrightarrowlnotforallxlnotphi)
4, 5brpi22<6([a/x]phitoexistsxphi)