Theorem fneq2 5305

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 2187	. . 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 5219	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5219	. 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 1353  dom cdm 4626  Fun wfun 5210   Fn wfn 5211

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 1447  ax-gen 1449  ax-4 1510  ax-17 1526  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-cleq 2170  df-fn 5219

This theorem is referenced by:  fneq2d  5307  fneq2i  5311  feq2  5349  foeq2  5435  f1o00  5496  eqfnfv2  5614  tfr0dm  6322  tfrlemisucaccv  6325  tfrlemi1  6332  tfrlemi14d  6333  tfrexlem  6334  tfr1onlemsucfn  6340  tfr1onlemsucaccv  6341  tfr1onlembxssdm  6343  tfr1onlembfn  6344  tfr1onlemaccex  6348  tfr1onlemres  6349  ixpeq1  6708  0fz1  10044