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Theorem fneq2 4931
 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 2046 . . 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 4848 . 2 (𝐹 Fn A ↔ (Fun 𝐹 dom 𝐹 = A))
4 df-fn 4848 . 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 1242  dom cdm 4288  Fun wfun 4839   Fn wfn 4840 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 1333  ax-gen 1335  ax-4 1397  ax-17 1416  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-cleq 2030  df-fn 4848 This theorem is referenced by:  fneq2d  4933  fneq2i  4937  feq2  4974  foeq2  5046  f1o00  5104  eqfnfv2  5209  tfr0  5878  tfrlemisucaccv  5880  tfrlemi1  5887  tfrlemi14d  5888  tfrexlem  5889  0fz1  8679
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