Theorem fneq2 5421

Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
fneq2	⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Proof of Theorem fneq2

Step	Hyp	Ref	Expression
1		eqeq2 2240	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 464	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 5331	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5331	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1397  dom cdm 4727  Fun wfun 5322   Fn wfn 5323

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1495  ax-gen 1497  ax-4 1558  ax-17 1574  ax-ext 2212

This theorem depends on definitions:  df-bi 117  df-cleq 2223  df-fn 5331

This theorem is referenced by:  fneq2d  5423  fneq2i  5427  feq2  5468  foeq2  5559  f1o00  5623  eqfnfv2  5748  tfr0dm  6493  tfrlemisucaccv  6496  tfrlemi1  6503  tfrlemi14d  6504  tfrexlem  6505  tfr1onlemsucfn  6511  tfr1onlemsucaccv  6512  tfr1onlembxssdm  6514  tfr1onlembfn  6515  tfr1onlemaccex  6519  tfr1onlemres  6520  ixpeq1  6883  0fz1  10285