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Theorem fneq2 5013
Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)
Assertion
Ref Expression
fneq2 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))

Proof of Theorem fneq2
StepHypRef Expression
1 eqeq2 2063 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 445 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 4930 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 4930 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 216 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102   = wceq 1257  dom cdm 4370  Fun wfun 4921   Fn wfn 4922
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 1350  ax-gen 1352  ax-4 1414  ax-17 1433  ax-ext 2036
This theorem depends on definitions:  df-bi 114  df-cleq 2047  df-fn 4930
This theorem is referenced by:  fneq2d  5015  fneq2i  5019  feq2  5056  foeq2  5128  f1o00  5186  eqfnfv2  5291  tfr0  5965  tfrlemisucaccv  5967  tfrlemi1  5974  tfrlemi14d  5975  tfrexlem  5976  0fz1  8981
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