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Theorem spim 1668
Description: Specialization, using implicit substitution. Compare Lemma 14 of [Tarski] p. 70. The spim 1668 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
StepHypRef Expression
1 spim.1 . . 3  |-  F/ x ps
21nfri 1453 . 2  |-  ( ps 
->  A. x ps )
3 spim.2 . 2  |-  ( x  =  y  ->  ( ph  ->  ps ) )
42, 3spimh 1667 1  |-  ( A. x ph  ->  ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1283   F/wnf 1390
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 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-nf 1391
This theorem is referenced by:  cbv3  1672  chvar  1682  spimv  1734  2spim  10837
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