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Theorem sbelx 2182
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 sbid2v 2181 . 2  |-  ( [ y  /  x ] [ x  /  y ] ph  <->  ph )
2 sb5 2157 . 2  |-  ( [ y  /  x ] [ x  /  y ] ph  <->  E. x ( x  =  y  /\  [
x  /  y ]
ph ) )
31, 2bitr3i 243 1  |-  ( ph  <->  E. x ( x  =  y  /\  [ x  /  y ] ph ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 177    /\ wa 359   E.wex 1547   [wsb 1655
This theorem is referenced by:  sbel2x  2183  pm13.196a  27490
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1662  ax-8 1683  ax-6 1740  ax-7 1745  ax-11 1757  ax-12 1946
This theorem depends on definitions:  df-bi 178  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656
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