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Theorem spimh 1698
Description: Specialization, using implicit substitition. Compare Lemma 14 of [Tarski] p. 70. The spim 1699 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 1414 . 2  |-  ( A. x ph  ->  A. x
( x  =  y  ->  A. x ps )
)
5 ax9o 1659 . 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 1312
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-4 1470  ax-i9 1493  ax-ial 1497
This theorem depends on definitions:  df-bi 116
This theorem is referenced by:  spim  1699
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