equf

Theorem.

Arguments:

a (coll), b (coll), x (sv),

Dummy variables:

y (sv),

Distinct variable conditions:

(y, a), (y, b), (y, x),

Hypotheses:

(x bound in b)
(x bound in a)

Assertions:

((aequivb)leftrightarrowforallx((xina)leftrightarrow(xinb)))

Proof:

Hyp Ref Line Expr
Hypo1(x bound in b)
2(x bound in a)
2df-bvt3(x bound in (yina))
1df-bvt4(x bound in (yinb))
3, 4bv-↔5(x bound in ((yina)leftrightarrow(yinb)))
df-equ6((aequivb)leftrightarrowforally((yina)leftrightarrow(yinb)))
ax17-bv7(y bound in ((xina)leftrightarrow(xinb)))
eqt<8((xeqy)to((xina)leftrightarrow(yina)))
eqt<9((xeqy)to((xinb)leftrightarrow(yinb)))
8, 9bi.bldbi<>d10((xeqy)to(((xina)leftrightarrow(xinb))leftrightarrow((yina)leftrightarrow(yinb))))
5, 7, 10cbval11(forallx((xina)leftrightarrow(xinb))leftrightarrowforally((yina)leftrightarrow(yinb)))
6, 11><bitr12((aequivb)leftrightarrowforallx((xina)leftrightarrow(xinb)))