Theorem fneq2 6625

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 2781	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 641	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6537	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6537	. 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 400   = wceq 1567  dom cdm 5659  Fun wfun 6528   Fn wfn 6529

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-cleq 2761  df-fn 6537

This theorem is referenced by:  fneq2d  6627  fneq2i  6631  feq2  6682  foeq2  6787  f1o00  6854  eqfnfv2  7024  frrlem1  8279  frrlem13  8291  tfrlem12  8372  ixpeq1  8902  ac5  10457  0fz1  13568  fconst7v  32902  esumcvgsum  34419  bnj90  35052  bnj919  35097  bnj535  35219  bnj1463  35384  fnchoice  45636