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Theorem sbequ8 2150
Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ8  |-  ( [ y  /  x ] ph 
<->  [ y  /  x ] ( x  =  y  ->  ph ) )

Proof of Theorem sbequ8
StepHypRef Expression
1 equsb1 2103 . . 3  |-  [ y  /  x ] x  =  y
21a1bi 329 . 2  |-  ( [ y  /  x ] ph 
<->  ( [ y  /  x ] x  =  y  ->  [ y  /  x ] ph ) )
3 sbim 2137 . 2  |-  ( [ y  /  x ]
( x  =  y  ->  ph )  <->  ( [
y  /  x ]
x  =  y  ->  [ y  /  x ] ph ) )
42, 3bitr4i 245 1  |-  ( [ y  /  x ] ph 
<->  [ y  /  x ] ( x  =  y  ->  ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 178   [wsb 1659
This theorem is referenced by:  sbie  2151
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-6 1745  ax-7 1750  ax-11 1762  ax-12 1951
This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-tru 1329  df-ex 1552  df-nf 1555  df-sb 1660
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