Theorem fneq2 6449

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 2748	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 632	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6361	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6361	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 317	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1543  dom cdm 5536  Fun wfun 6352   Fn wfn 6353

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-9 2122  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1788  df-cleq 2728  df-fn 6361

This theorem is referenced by:  fneq2d  6451  fneq2i  6455  feq2  6505  foeq2  6608  f1o00  6673  eqfnfv2  6831  frrlem1  8005  frrlem13  8017  wfrlem1  8032  wfrlem15  8047  tfrlem12  8103  ixpeq1  8567  ac5  10056  0fz1  13097  esumcvgsum  31722  bnj90  32367  bnj919  32413  bnj535  32537  bnj1463  32702  fnchoice  42186