Theorem fneq2 5426

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 2241	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 464	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 5336	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5336	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1398  dom cdm 4731  Fun wfun 5327   Fn wfn 5328

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1496  ax-gen 1498  ax-4 1559  ax-17 1575  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-fn 5336

This theorem is referenced by:  fneq2d  5428  fneq2i  5432  feq2  5473  foeq2  5565  f1o00  5629  eqfnfv2  5754  tfr0dm  6531  tfrlemisucaccv  6534  tfrlemi1  6541  tfrlemi14d  6542  tfrexlem  6543  tfr1onlemsucfn  6549  tfr1onlemsucaccv  6550  tfr1onlembxssdm  6552  tfr1onlembfn  6553  tfr1onlemaccex  6557  tfr1onlemres  6558  ixpeq1  6921  0fz1  10325