eqid

Theorem.

Arguments:

x (st),

Dummy variables:

y (sv),

Distinct variable conditions:

(y, x),

Assertions:

(xeqx)

Proof:

Hyp Ref Line Expr
ax9ex1existsy(yeqx)
ax172((xeqx)toforally(xeqx))
2df-Bvp3(y bound in (xeqx))
ax84((yeqx)to((yeqx)to(xeqx)))
4imabs5((yeqx)to(xeqx))
3, 5imgen<i6(existsy(yeqx)to(xeqx))
1, 6ax-mp7(xeqx)