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Theorem fneq2 5089
Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)
Assertion
Ref Expression
fneq2  |-  ( A  =  B  ->  ( F  Fn  A  <->  F  Fn  B ) )

Proof of Theorem fneq2
StepHypRef Expression
1 eqeq2 2097 . . 3  |-  ( A  =  B  ->  ( dom  F  =  A  <->  dom  F  =  B ) )
21anbi2d 452 . 2  |-  ( A  =  B  ->  (
( Fun  F  /\  dom  F  =  A )  <-> 
( Fun  F  /\  dom  F  =  B ) ) )
3 df-fn 5005 . 2  |-  ( F  Fn  A  <->  ( Fun  F  /\  dom  F  =  A ) )
4 df-fn 5005 . 2  |-  ( F  Fn  B  <->  ( Fun  F  /\  dom  F  =  B ) )
52, 3, 43bitr4g 221 1  |-  ( A  =  B  ->  ( F  Fn  A  <->  F  Fn  B ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103    = wceq 1289   dom cdm 4428   Fun wfun 4996    Fn wfn 4997
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 1381  ax-gen 1383  ax-4 1445  ax-17 1464  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-cleq 2081  df-fn 5005
This theorem is referenced by:  fneq2d  5091  fneq2i  5095  feq2  5132  foeq2  5214  f1o00  5272  eqfnfv2  5382  tfr0dm  6069  tfrlemisucaccv  6072  tfrlemi1  6079  tfrlemi14d  6080  tfrexlem  6081  tfr1onlemsucfn  6087  tfr1onlemsucaccv  6088  tfr1onlembxssdm  6090  tfr1onlembfn  6091  tfr1onlemaccex  6095  tfr1onlemres  6096  0fz1  9428
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