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Theorem spimh 1667
Description: Specialization, using implicit substitition. 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 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
StepHypRef Expression
1 spimh.2 . . . 4  |-  ( x  =  y  ->  ( ph  ->  ps ) )
2 spimh.1 . . . 4  |-  ( ps 
->  A. x ps )
31, 2syl6com 35 . . 3  |-  ( ph  ->  ( x  =  y  ->  A. x ps )
)
43alimi 1385 . 2  |-  ( A. x ph  ->  A. x
( x  =  y  ->  A. x ps )
)
5 ax9o 1629 . 2  |-  ( A. x ( x  =  y  ->  A. x ps )  ->  ps )
64, 5syl 14 1  |-  ( A. x ph  ->  ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1283
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
This theorem is referenced by:  spim  1668
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