Theorem fneq2 5372

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 2216	. . 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 5283	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5283	. 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 1373  dom cdm 4683  Fun wfun 5274   Fn wfn 5275

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 1471  ax-gen 1473  ax-4 1534  ax-17 1550  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-cleq 2199  df-fn 5283

This theorem is referenced by:  fneq2d  5374  fneq2i  5378  feq2  5419  foeq2  5507  f1o00  5570  eqfnfv2  5691  tfr0dm  6421  tfrlemisucaccv  6424  tfrlemi1  6431  tfrlemi14d  6432  tfrexlem  6433  tfr1onlemsucfn  6439  tfr1onlemsucaccv  6440  tfr1onlembxssdm  6442  tfr1onlembfn  6443  tfr1onlemaccex  6447  tfr1onlemres  6448  ixpeq1  6809  0fz1  10187