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Theorem sbequ1 2097
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 pm3.4 570 . . 3  |-  ( ( x  =  y  /\  ph )  ->  ( x  =  y  ->  ph )
)
2 19.8a 1955 . . 3  |-  ( ( x  =  y  /\  ph )  ->  E. x
( x  =  y  /\  ph ) )
3 df-sb 1806 . . 3  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
41, 2, 3sylanbrc 677 . 2  |-  ( ( x  =  y  /\  ph )  ->  [ y  /  x ] ph )
54ex 441 1  |-  ( x  =  y  ->  ( ph  ->  [ y  /  x ] ph ) )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    /\ wa 376   E.wex 1671   [wsb 1805
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-12 1950
This theorem depends on definitions:  df-bi 190  df-an 378  df-ex 1672  df-sb 1806
This theorem is referenced by:  sbequ12  2098  dfsb2  2222  sbequi  2224  sbi1  2241  2eu6  2407  sb5ALT  36952  2pm13.193  36989  2pm13.193VD  37363  sb5ALTVD  37373
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