Theorem fneq2 6574

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 2741	. . 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 6485	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6485	. 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 1540  dom cdm 5619  Fun wfun 6476   Fn wfn 6477

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-cleq 2721  df-fn 6485

This theorem is referenced by:  fneq2d  6576  fneq2i  6580  feq2  6631  foeq2  6733  f1o00  6799  eqfnfv2  6966  frrlem1  8219  frrlem13  8231  tfrlem12  8311  ixpeq1  8835  ac5  10371  0fz1  13447  fconst7v  32565  esumcvgsum  34055  bnj90  34689  bnj919  34734  bnj535  34857  bnj1463  35022  fnchoice  45007