naleqcoms

Theorem.

Arguments:

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

Hypotheses:

(lnotforallx(xeqy)tophi)

Assertions:

(lnotforally(yeqx)tophi)

Proof:

Hyp Ref Line Expr
Hypo1(lnotforallx(xeqy)tophi)
aleqcom2(forallx(xeqy)toforally(yeqx))
1con<3(lnotphitoforallx(xeqy))
2, 3syl4(lnotphitoforally(yeqx))
4con<5(lnotforally(yeqx)tophi)