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Theorem sbequ2 1699
Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbequ2  |-  ( x  =  y  ->  ( [ y  /  x ] ph  ->  ph ) )

Proof of Theorem sbequ2
StepHypRef Expression
1 df-sb 1693 . 2  |-  ( [ y  /  x ] ph 
<->  ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
) )
2 simpl 107 . . 3  |-  ( ( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
)  ->  ( x  =  y  ->  ph )
)
32com12 30 . 2  |-  ( x  =  y  ->  (
( ( x  =  y  ->  ph )  /\  E. x ( x  =  y  /\  ph )
)  ->  ph ) )
41, 3syl5bi 150 1  |-  ( x  =  y  ->  ( [ y  /  x ] ph  ->  ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102   E.wex 1426   [wsb 1692
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104
This theorem depends on definitions:  df-bi 115  df-sb 1693
This theorem is referenced by:  stdpc7  1700  sbequ12  1701  sbequi  1767  mo23  1989  mopick  2026
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