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Theorem sbequ1 1695
Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )

Proof of Theorem sbequ1
StepHypRef Expression
1 pm3.4 326 . . 3  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  ->  ph )
)
2 19.8a 1525 . . 3  |-  ( ( x  =  y  /\  ph )  ->  E. x
( x  =  y  /\  ph ) )
3 df-sb 1690 . . 3  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
41, 2, 3sylanbrc 408 . 2  |-  ( ( x  =  y  /\  ph )  ->  [ y  /  x ] ph )
54ex 113 1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102   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-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  sbequ12  1698  sbequi  1764  sb6rf  1778  mo2n  1973  bj-bdfindes  11332  bj-findes  11364
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