Theorem spimh 1665

Description: Specialization, using implicit substitition. Compare Lemma 14 of [Tarski] p. 70. The spim 1666 series of theorems requires that only one direction of the substitution hypothesis hold. (Contributed by NM, 5-Aug-1993.) (Revised by NM, 8-May-2008.) (New usage is discouraged.)

Hypotheses

Ref	Expression
spimh.1	|- ( ps -> A. x ps )
spimh.2	|- ( x = y -> ( ph -> ps ) )

Assertion

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

Proof of Theorem spimh

Step	Hyp	Ref	Expression
1		spimh.2	. . . 4 |- ( x = y -> ( ph -> ps ) )
2		spimh.1	. . . 4 |- ( ps -> A. x ps )
3	1, 2	syl6com 35	. . 3 |- ( ph -> ( x = y -> A. x ps ) )
4	3	alimi 1384	. 2 |- ( A. x ph -> A. x ( x = y -> A. x ps ) )
5		ax9o 1628	. 2 |- ( A. x ( x = y -> A. x ps ) -> ps )
6	4, 5	syl 14	1 |- ( A. x ph -> ps )

Syntax hints:   -> wi 4   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-ie1 1422  ax-ie2 1423  ax-4 1440  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  spim  1666