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Theorem spim 1975
Description: Specialization, using implicit substitution. Compare Lemma 14 of [Tarski] p. 70. The spim 1975 series of theorems requires that only one direction of the substitution hypothesis hold. (Contributed by NM, 10-Jan-1993.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof shortened by Wolf Lammen, 18-Feb-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 . 2  |-  F/ x ps
2 ax6e 1971 . . 3  |-  E. x  x  =  y
3 spim.2 . . 3  |-  ( x  =  y  ->  ( ph  ->  ps ) )
42, 3eximii 1637 . 2  |-  E. x
( ph  ->  ps )
51, 419.36i 1914 1  |-  ( A. x ph  ->  ps )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4   A.wal 1377   F/wnf 1599
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1601  ax-4 1612  ax-5 1680  ax-6 1719  ax-7 1739  ax-10 1786  ax-12 1803  ax-13 1968
This theorem depends on definitions:  df-bi 185  df-an 371  df-ex 1597  df-nf 1600
This theorem is referenced by:  spimv  1978  chvar  1982  cbv3  1984
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