Theorem fneq2 5320

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 2199	. . 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 5234	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5234	. 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 1364  dom cdm 4641  Fun wfun 5225   Fn wfn 5226

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 1458  ax-gen 1460  ax-4 1521  ax-17 1537  ax-ext 2171

This theorem depends on definitions:  df-bi 117  df-cleq 2182  df-fn 5234

This theorem is referenced by:  fneq2d  5322  fneq2i  5326  feq2  5364  foeq2  5450  f1o00  5511  eqfnfv2  5630  tfr0dm  6341  tfrlemisucaccv  6344  tfrlemi1  6351  tfrlemi14d  6352  tfrexlem  6353  tfr1onlemsucfn  6359  tfr1onlemsucaccv  6360  tfr1onlembxssdm  6362  tfr1onlembfn  6363  tfr1onlemaccex  6367  tfr1onlemres  6368  ixpeq1  6727  0fz1  10063