Theorem hbal 1406

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 1384	. 2 |- ( A. y ph -> A. y A. x ph )
3		ax-7 1377	. 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 1282

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-5 1376  ax-7 1377  ax-gen 1378

This theorem is referenced by:  hba2  1483  nfal  1508  aaanh  1518  hbex  1567  pm11.53  1816  euf  1946  hbral  2395