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Theorem spim 1915
Description: Specialization, using implicit substitution. Compare Lemma 14 of [Tarski] p. 70. The spim 1915 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.)
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 spim.2 . . 3  |-  ( x  =  y  ->  ( ph  ->  ps ) )
32ax-gen 1533 . 2  |-  A. x
( x  =  y  ->  ( ph  ->  ps ) )
4 spimt 1914 . 2  |-  ( ( F/ x ps  /\  A. x ( x  =  y  ->  ( ph  ->  ps ) ) )  ->  ( A. x ph  ->  ps ) )
51, 3, 4mp2an 653 1  |-  ( A. x ph  ->  ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1527   F/wnf 1531
This theorem is referenced by:  spime  1916  chvar  1926  spimv  1930
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532
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