Theorem fneq2 5357

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 5271	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5271	. 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 4673  Fun wfun 5262   Fn wfn 5263

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 5271

This theorem is referenced by:  fneq2d  5359  fneq2i  5363  feq2  5403  foeq2  5489  f1o00  5551  eqfnfv2  5672  tfr0dm  6398  tfrlemisucaccv  6401  tfrlemi1  6408  tfrlemi14d  6409  tfrexlem  6410  tfr1onlemsucfn  6416  tfr1onlemsucaccv  6417  tfr1onlembxssdm  6419  tfr1onlembfn  6420  tfr1onlemaccex  6424  tfr1onlemres  6425  ixpeq1  6786  0fz1  10149