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Theorem fneq2 6594
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 2748 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 629 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 6499 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 6499 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 313 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396   = wceq 1541  dom cdm 5633  Fun wfun 6490   Fn wfn 6491
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-9 2116  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1782  df-cleq 2728  df-fn 6499
This theorem is referenced by:  fneq2d  6596  fneq2i  6600  feq2  6650  foeq2  6753  f1o00  6819  eqfnfv2  6983  frrlem1  8217  frrlem13  8229  wfrlem1OLD  8254  wfrlem15OLD  8269  tfrlem12  8335  ixpeq1  8846  ac5  10413  0fz1  13461  esumcvgsum  32687  bnj90  33334  bnj919  33379  bnj535  33502  bnj1463  33667  fnchoice  43224
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