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Theorem sbc5 2905
Description: An equivalence for class substitution. (Contributed by NM, 23-Aug-1993.) (Revised by Mario Carneiro, 12-Oct-2016.)
Assertion
Ref Expression
sbc5  |-  ( [. A  /  x ]. ph  <->  E. x
( x  =  A  /\  ph ) )
Distinct variable group:    x, A
Allowed substitution hint:    ph( x)

Proof of Theorem sbc5
Dummy variable  y is distinct from all other variables.
StepHypRef Expression
1 sbcex 2890 . 2  |-  ( [. A  /  x ]. ph  ->  A  e.  _V )
2 exsimpl 1581 . . 3  |-  ( E. x ( x  =  A  /\  ph )  ->  E. x  x  =  A )
3 isset 2666 . . 3  |-  ( A  e.  _V  <->  E. x  x  =  A )
42, 3sylibr 133 . 2  |-  ( E. x ( x  =  A  /\  ph )  ->  A  e.  _V )
5 dfsbcq2 2885 . . 3  |-  ( y  =  A  ->  ( [ y  /  x ] ph  <->  [. A  /  x ]. ph ) )
6 eqeq2 2127 . . . . 5  |-  ( y  =  A  ->  (
x  =  y  <->  x  =  A ) )
76anbi1d 460 . . . 4  |-  ( y  =  A  ->  (
( x  =  y  /\  ph )  <->  ( x  =  A  /\  ph )
) )
87exbidv 1781 . . 3  |-  ( y  =  A  ->  ( E. x ( x  =  y  /\  ph )  <->  E. x ( x  =  A  /\  ph )
) )
9 sb5 1843 . . 3  |-  ( [ y  /  x ] ph 
<->  E. x ( x  =  y  /\  ph ) )
105, 8, 9vtoclbg 2721 . 2  |-  ( A  e.  _V  ->  ( [. A  /  x ]. ph  <->  E. x ( x  =  A  /\  ph ) ) )
111, 4, 10pm5.21nii 678 1  |-  ( [. A  /  x ]. ph  <->  E. x
( x  =  A  /\  ph ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 103    <-> wb 104    = wceq 1316   E.wex 1453    e. wcel 1465   [wsb 1720   _Vcvv 2660   [.wsbc 2882
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-v 2662  df-sbc 2883
This theorem is referenced by:  sbc6g  2906  sbc7  2908  sbciegft  2911  sbccomlem  2955  csb2  2977  rexsns  3533
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