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Theorem sbelx 1915
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 1460 . 2  |-  ( ph  ->  A. x ph )
21sb5rf 1774 1  |-  ( ph  <->  E. x ( x  =  y  /\  [ x  /  y ] ph ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 102    <-> wb 103   E.wex 1422   [wsb 1686
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-11 1438  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-sb 1687
This theorem is referenced by:  sbel2x  1916
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