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Theorem sb6x 1779
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)
Hypothesis
Ref Expression
sb6x.1  |-  ( ph  ->  A. x ph )
Assertion
Ref Expression
sb6x  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3  |-  ( ph  ->  A. x ph )
21sbh 1776 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 172 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsalh 1726 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  ph )
52, 4bitr4i 187 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 105   A.wal 1351   [wsb 1762
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-4 1510  ax-i9 1530  ax-ial 1534
This theorem depends on definitions:  df-bi 117  df-sb 1763
This theorem is referenced by: (None)
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