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Theorem sbelx 2009
Description: Elimination of substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbelx  |-  ( ph  <->  E. x ( x  =  y  /\  [ x  /  y ] ph ) )
Distinct variable groups:    x, y    ph, x
Allowed substitution hint:    ph( y)

Proof of Theorem sbelx
StepHypRef Expression
1 ax-17 1537 . 2  |-  ( ph  ->  A. x ph )
21sb5rf 1863 1  |-  ( ph  <->  E. x ( x  =  y  /\  [ x  /  y ] ph ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 104    <-> wb 105   E.wex 1503   [wsb 1773
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545
This theorem depends on definitions:  df-bi 117  df-sb 1774
This theorem is referenced by:  sbel2x  2010
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