Theorem spim 1666

Description: Specialization, using implicit substitution. 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 Mario Carneiro, 3-Oct-2016.) (Proof rewritten by Jim Kingdon, 10-Jun-2018.)

Hypotheses

Ref	Expression
spim.1	|- F/ x ps
spim.2	|- ( x = y -> ( ph -> ps ) )

Assertion

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

Proof of Theorem spim

Step	Hyp	Ref	Expression
1		spim.1	. . 3 |- F/ x ps
2	1	nfri 1452	. 2 |- ( ps -> A. x ps )
3		spim.2	. 2 |- ( x = y -> ( ph -> ps ) )
4	2, 3	spimh 1665	1 |- ( A. x ph -> ps )

Syntax hints:   -> wi 4   A.wal 1282   F/wnf 1389

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  df-nf 1390

This theorem is referenced by:  cbv3  1670  chvar  1680  spimv  1732  2spim  10577