Theorem fneq2 5445

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 2242	. . 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 5355	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5355	. 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 4749  Fun wfun 5346   Fn wfn 5347

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 2214

This theorem depends on definitions:  df-bi 117  df-cleq 2225  df-fn 5355

This theorem is referenced by:  fneq2d  5447  fneq2i  5451  feq2  5492  foeq2  5587  f1o00  5651  eqfnfv2  5776  tfr0dm  6553  tfrlemisucaccv  6556  tfrlemi1  6563  tfrlemi14d  6564  tfrexlem  6565  tfr1onlemsucfn  6571  tfr1onlemsucaccv  6572  tfr1onlembxssdm  6574  tfr1onlembfn  6575  tfr1onlemaccex  6579  tfr1onlemres  6580  ixpeq1  6944  0fz1  10379