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Theorem sbequ1 2009
Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.)
Assertion
Ref Expression
sbequ1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )

Proof of Theorem sbequ1
StepHypRef Expression
1 pm3.4 559 . . 3  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  ->  ph )
)
2 19.8a 1875 . . 3  |-  ( ( x  =  y  /\  ph )  ->  E. x
( x  =  y  /\  ph ) )
3 df-sb 1758 . . 3  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
41, 2, 3sylanbrc 662 . 2  |-  ( ( x  =  y  /\  ph )  ->  [ y  /  x ] ph )
54ex 432 1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    /\ wa 367   E.wex 1627   [wsb 1757
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1633  ax-4 1646  ax-5 1719  ax-6 1765  ax-7 1808  ax-12 1872
This theorem depends on definitions:  df-bi 185  df-an 369  df-ex 1628  df-sb 1758
This theorem is referenced by:  sbequ12  2010  dfsb2  2128  sbequi  2130  sbi1  2147  2eu6  2324  sb5ALT  33663  2pm13.193  33700  2pm13.193VD  34085  sb5ALTVD  34095
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