Theorem fneq2 6629

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 2773	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 642	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6540	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6540	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 317	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  dom cdm 5651  Fun wfun 6531   Fn wfn 6532

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2753  df-fn 6540

This theorem is used by:  fneq2d  6631  fneq2i  6635  feq2  6686  foeq2  6791  f1o00  6858  eqfnfv2  7028  frrlem1  8297  frrlem13  8309  tfrlem12  8390  ixpeq1  8929  ac5  10548  0fz1  13670  fconst7v  33207  esumcvgsum  34713  bnj90  35346  bnj919  35391  bnj535  35513  bnj1463  35678  fnchoice  46015