Theorem fneq2 6584

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 630	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6495	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6495	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 314	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1541  dom cdm 5624  Fun wfun 6486   Fn wfn 6487

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-9 2123  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-cleq 2728  df-fn 6495

This theorem is referenced by:  fneq2d  6586  fneq2i  6590  feq2  6641  foeq2  6743  f1o00  6809  eqfnfv2  6977  frrlem1  8228  frrlem13  8240  tfrlem12  8320  ixpeq1  8846  ac5  10387  0fz1  13460  fconst7v  32698  esumcvgsum  34245  bnj90  34878  bnj919  34923  bnj535  35046  bnj1463  35211  fnchoice  45274