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Theorem sbequ1 1859
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 pm3.4 544 . . 3  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  ->  ph )
)
2 19.8a 1718 . . 3  |-  ( ( x  =  y  /\  ph )  ->  E. x
( x  =  y  /\  ph ) )
3 df-sb 1630 . . 3  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
41, 2, 3sylanbrc 645 . 2  |-  ( ( x  =  y  /\  ph )  ->  [ y  /  x ] ph )
54ex 423 1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 358   E.wex 1528   [wsb 1629
This theorem is referenced by:  sbequ12  1860  dfsb2  1995  sbequi  1999  sbn  2002  sbi1  2003  sb6rf  2031  mo  2165  sb5ALT  28288  2pm13.193  28318  2pm13.193VD  28679  sb5ALTVD  28689
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-11 1715
This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1529  df-sb 1630
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