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Theorem fneq2 4988
 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 2049 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 437 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 4905 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 4905 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 212 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  dom cdm 4345  Fun wfun 4896   Fn wfn 4897 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400  ax-17 1419  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-cleq 2033  df-fn 4905 This theorem is referenced by:  fneq2d  4990  fneq2i  4994  feq2  5031  foeq2  5103  f1o00  5161  eqfnfv2  5266  tfr0  5937  tfrlemisucaccv  5939  tfrlemi1  5946  tfrlemi14d  5947  tfrexlem  5948  0fz1  8909
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