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Theorem sbequ8 1772
Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.) (Proof revised by Jim Kingdon, 20-Jan-2018.)
Assertion
Ref Expression
sbequ8  |-  ( [ y  /  x ] ph 
<->  [ y  /  x ] ( x  =  y  ->  ph ) )

Proof of Theorem sbequ8
StepHypRef Expression
1 pm5.4 247 . . 3  |-  ( ( x  =  y  -> 
( x  =  y  ->  ph ) )  <->  ( x  =  y  ->  ph )
)
2 simpl 107 . . . . . 6  |-  ( ( x  =  y  /\  ( x  =  y  ->  ph ) )  ->  x  =  y )
3 pm3.35 339 . . . . . 6  |-  ( ( x  =  y  /\  ( x  =  y  ->  ph ) )  ->  ph )
42, 3jca 300 . . . . 5  |-  ( ( x  =  y  /\  ( x  =  y  ->  ph ) )  -> 
( x  =  y  /\  ph ) )
5 simpl 107 . . . . . 6  |-  ( ( x  =  y  /\  ph )  ->  x  =  y )
6 pm3.4 326 . . . . . 6  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  ->  ph )
)
75, 6jca 300 . . . . 5  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  /\  (
x  =  y  ->  ph ) ) )
84, 7impbii 124 . . . 4  |-  ( ( x  =  y  /\  ( x  =  y  ->  ph ) )  <->  ( x  =  y  /\  ph )
)
98exbii 1539 . . 3  |-  ( E. x ( x  =  y  /\  ( x  =  y  ->  ph )
)  <->  E. x ( x  =  y  /\  ph ) )
101, 9anbi12i 448 . 2  |-  ( ( ( x  =  y  ->  ( x  =  y  ->  ph ) )  /\  E. x ( x  =  y  /\  ( x  =  y  ->  ph ) ) )  <-> 
( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
11 df-sb 1690 . 2  |-  ( [ y  /  x ]
( x  =  y  ->  ph )  <->  ( (
x  =  y  -> 
( x  =  y  ->  ph ) )  /\  E. x ( x  =  y  /\  ( x  =  y  ->  ph )
) ) )
12 df-sb 1690 . 2  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
1310, 11, 123bitr4ri 211 1  |-  ( [ y  /  x ] ph 
<->  [ y  /  x ] ( x  =  y  ->  ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103   E.wex 1424   [wsb 1689
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1379  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443  ax-ial 1470
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  sbidm  1776
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