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Theorem fneq2 6438
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 2830 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 628 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 6351 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 6351 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 315 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396   = wceq 1528  dom cdm 5548  Fun wfun 6342   Fn wfn 6343
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-9 2115  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1772  df-cleq 2811  df-fn 6351
This theorem is referenced by:  fneq2d  6440  fneq2i  6444  feq2  6489  foeq2  6580  f1o00  6642  eqfnfv2  6795  wfrlem1  7943  wfrlem15  7958  tfrlem12  8014  ixpeq1  8460  ac5  9887  0fz1  12915  esumcvgsum  31246  bnj90  31891  bnj919  31937  bnj535  32061  bnj1463  32224  frrlem1  33020  frrlem13  33032  fnchoice  41163
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