Theorem fneq2 5362

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 2214	. . 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 5273	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5273	. 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 1372  dom cdm 4674  Fun wfun 5264   Fn wfn 5265

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 1469  ax-gen 1471  ax-4 1532  ax-17 1548  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-cleq 2197  df-fn 5273

This theorem is referenced by:  fneq2d  5364  fneq2i  5368  feq2  5408  foeq2  5494  f1o00  5556  eqfnfv2  5677  tfr0dm  6407  tfrlemisucaccv  6410  tfrlemi1  6417  tfrlemi14d  6418  tfrexlem  6419  tfr1onlemsucfn  6425  tfr1onlemsucaccv  6426  tfr1onlembxssdm  6428  tfr1onlembfn  6429  tfr1onlemaccex  6433  tfr1onlemres  6434  ixpeq1  6795  0fz1  10166