Theorem hbal 1500

Description: If x is not free in ph, it is not free in A. y ph. (Contributed by NM, 5-Aug-1993.)

Hypothesis

Ref	Expression
hbal.1	|- ( ph -> A. x ph )

Assertion

Ref	Expression
hbal	|- ( A. y ph -> A. x A. y ph )

Proof of Theorem hbal

Step	Hyp	Ref	Expression
1		hbal.1	. . 3 |- ( ph -> A. x ph )
2	1	alimi 1478	. 2 |- ( A. y ph -> A. y A. x ph )
3		ax-7 1471	. 2 |- ( A. y A. x ph -> A. x A. y ph )
4	2, 3	syl 14	1 |- ( A. y ph -> A. x A. y ph )

Syntax hints:   -> wi 4   A.wal 1371

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-5 1470  ax-7 1471  ax-gen 1472

This theorem is referenced by:  hba2  1574  aaanh  1609  hbex  1659  pm11.53  1919  euf  2059  hbral  2535