Theorem hbbi 1480

Description: If x is not free in ph and ps, it is not free in ( ph <-> ps ). (Contributed by NM, 5-Aug-1993.)

Hypotheses

Ref	Expression
hb.1	|- ( ph -> A. x ph )
hb.2	|- ( ps -> A. x ps )

Assertion

Ref	Expression
hbbi	|- ( ( ph <-> ps ) -> A. x ( ph <-> ps ) )

Proof of Theorem hbbi

Step	Hyp	Ref	Expression
1		dfbi2 380	. 2 |- ( ( ph <-> ps ) <-> ( ( ph -> ps ) /\ ( ps -> ph ) ) )
2		hb.1	. . . 4 |- ( ph -> A. x ph )
3		hb.2	. . . 4 |- ( ps -> A. x ps )
4	2, 3	hbim 1477	. . 3 |- ( ( ph -> ps ) -> A. x ( ph -> ps ) )
5	3, 2	hbim 1477	. . 3 |- ( ( ps -> ph ) -> A. x ( ps -> ph ) )
6	4, 5	hban 1479	. 2 |- ( ( ( ph -> ps ) /\ ( ps -> ph ) ) -> A. x ( ( ph -> ps ) /\ ( ps -> ph ) ) )
7	1, 6	hbxfrbi 1401	1 |- ( ( ph <-> ps ) -> A. x ( ph <-> ps ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103   A.wal 1282

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-4 1440  ax-i5r 1468

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  euf  1946  sb8euh  1964