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Theorem fneq2 6018
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 2662 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 740 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 5929 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 5929 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 303 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 383   = wceq 1523  dom cdm 5143  Fun wfun 5920   Fn wfn 5921
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-an 385  df-ex 1745  df-cleq 2644  df-fn 5929
This theorem is referenced by:  fneq2d  6020  fneq2i  6024  feq2  6065  foeq2  6150  f1o00  6209  eqfnfv2  6352  wfrlem1  7459  wfrlem15  7474  tfrlem12  7530  ixpeq1  7961  ac5  9337  0fz1  12399  esumcvgsum  30278  bnj90  30916  bnj919  30963  bnj535  31086  bnj1463  31249  frrlem1  31905  fnchoice  39502
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