aleqcom

Theorem.

Arguments:

x (sv), y (sv),

Assertions:

(forallx(xeqy)toforally(yeqx))

Proof:

Hyp Ref Line Expr
ax101(forallx(xeqy)to(forallx(xeqy)toforally(xeqy)))
1imabs2(forallx(xeqy)toforally(xeqy))
eqcomi3((xeqy)to(yeqx))
3im.bldal4(forally(xeqy)toforally(yeqx))
2, 4syl5(forallx(xeqy)toforally(yeqx))