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Theorem eqrdav 2055
Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008.)
Hypotheses
Ref Expression
eqrdav.1  |-  ( (
ph  /\  x  e.  A )  ->  x  e.  C )
eqrdav.2  |-  ( (
ph  /\  x  e.  B )  ->  x  e.  C )
eqrdav.3  |-  ( (
ph  /\  x  e.  C )  ->  (
x  e.  A  <->  x  e.  B ) )
Assertion
Ref Expression
eqrdav  |-  ( ph  ->  A  =  B )
Distinct variable groups:    x, A    x, B    ph, x
Allowed substitution hint:    C( x)

Proof of Theorem eqrdav
StepHypRef Expression
1 eqrdav.1 . . . 4  |-  ( (
ph  /\  x  e.  A )  ->  x  e.  C )
2 eqrdav.3 . . . . . 6  |-  ( (
ph  /\  x  e.  C )  ->  (
x  e.  A  <->  x  e.  B ) )
32biimpd 136 . . . . 5  |-  ( (
ph  /\  x  e.  C )  ->  (
x  e.  A  ->  x  e.  B )
)
43impancom 251 . . . 4  |-  ( (
ph  /\  x  e.  A )  ->  (
x  e.  C  ->  x  e.  B )
)
51, 4mpd 13 . . 3  |-  ( (
ph  /\  x  e.  A )  ->  x  e.  B )
6 eqrdav.2 . . . 4  |-  ( (
ph  /\  x  e.  B )  ->  x  e.  C )
72exbiri 368 . . . . . 6  |-  ( ph  ->  ( x  e.  C  ->  ( x  e.  B  ->  x  e.  A ) ) )
87com23 76 . . . . 5  |-  ( ph  ->  ( x  e.  B  ->  ( x  e.  C  ->  x  e.  A ) ) )
98imp 119 . . . 4  |-  ( (
ph  /\  x  e.  B )  ->  (
x  e.  C  ->  x  e.  A )
)
106, 9mpd 13 . . 3  |-  ( (
ph  /\  x  e.  B )  ->  x  e.  A )
115, 10impbida 538 . 2  |-  ( ph  ->  ( x  e.  A  <->  x  e.  B ) )
1211eqrdv 2054 1  |-  ( ph  ->  A  =  B )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 101    <-> wb 102    = wceq 1259    e. wcel 1409
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-17 1435  ax-ext 2038
This theorem depends on definitions:  df-bi 114  df-cleq 2049
This theorem is referenced by:  fzdifsuc  9045
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