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Theorem sbelx 2063
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 2062 . 2  |-  ( [ y  /  x ] [ x  /  y ] ph  <->  ph )
2 sb5 2039 . 2  |-  ( [ y  /  x ] [ x  /  y ] ph  <->  E. x ( x  =  y  /\  [
x  /  y ]
ph ) )
31, 2bitr3i 242 1  |-  ( ph  <->  E. x ( x  =  y  /\  [ x  /  y ] ph ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 176    /\ wa 358   E.wex 1528   [wsb 1629
This theorem is referenced by:  sbel2x  2064  pm13.196a  27614
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630
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