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Theorem fneq2 6419
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 2813 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 631 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 6331 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 6331 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 317 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 399   = wceq 1538  dom cdm 5523  Fun wfun 6322   Fn wfn 6323
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 1911  ax-6 1970  ax-7 2015  ax-9 2122  ax-ext 2773
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-cleq 2794  df-fn 6331
This theorem is referenced by:  fneq2d  6421  fneq2i  6425  feq2  6473  foeq2  6566  f1o00  6628  eqfnfv2  6784  wfrlem1  7941  wfrlem15  7956  tfrlem12  8012  ixpeq1  8459  ac5  9892  0fz1  12926  esumcvgsum  31461  bnj90  32106  bnj919  32152  bnj535  32276  bnj1463  32441  frrlem1  33237  frrlem13  33249  fnchoice  41655
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