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Theorem sb6x 2082
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sb6x.1  |-  F/ x ph
Assertion
Ref Expression
sb6x  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3  |-  F/ x ph
21sbf 2078 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 237 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsal 1993 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  ph )
52, 4bitr4i 252 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    <-> wb 184   A.wal 1368   F/wnf 1590   [wsb 1702
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1592  ax-4 1603  ax-5 1671  ax-6 1710  ax-7 1730  ax-10 1777  ax-12 1794  ax-13 1952
This theorem depends on definitions:  df-bi 185  df-an 371  df-ex 1588  df-nf 1591  df-sb 1703
This theorem is referenced by: (None)
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