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Theorem sb6x 1982
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-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 1979 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 228 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsal 1913 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  ph )
52, 4bitr4i 243 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176   A.wal 1530   F/wnf 1534   [wsb 1638
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878
This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1532  df-nf 1535  df-sb 1639
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