Theorem fneq2 5419

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 2241	. . 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 5329	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5329	. 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 4725  Fun wfun 5320   Fn wfn 5321

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 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-fn 5329

This theorem is referenced by:  fneq2d  5421  fneq2i  5425  feq2  5466  foeq2  5556  f1o00  5620  eqfnfv2  5745  tfr0dm  6488  tfrlemisucaccv  6491  tfrlemi1  6498  tfrlemi14d  6499  tfrexlem  6500  tfr1onlemsucfn  6506  tfr1onlemsucaccv  6507  tfr1onlembxssdm  6509  tfr1onlembfn  6510  tfr1onlemaccex  6514  tfr1onlemres  6515  ixpeq1  6878  0fz1  10280