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Theorem sb6x 2127
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 2123 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 237 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsal 2040 . 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 1396   F/wnf 1621   [wsb 1744
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1623  ax-4 1636  ax-5 1709  ax-6 1752  ax-7 1795  ax-10 1842  ax-12 1859  ax-13 2004
This theorem depends on definitions:  df-bi 185  df-an 369  df-ex 1618  df-nf 1622  df-sb 1745
This theorem is referenced by: (None)
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