Theorem fneq2 4931

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 2046	. . 3 ⊢ (A = B → (dom 𝐹 = A ↔ dom 𝐹 = B))
2	1	anbi2d 437	. 2 ⊢ (A = B → ((Fun 𝐹 ∧ dom 𝐹 = A) ↔ (Fun 𝐹 ∧ dom 𝐹 = B)))
3		df-fn 4848	. 2 ⊢ (𝐹 Fn A ↔ (Fun 𝐹 ∧ dom 𝐹 = A))
4		df-fn 4848	. 2 ⊢ (𝐹 Fn B ↔ (Fun 𝐹 ∧ dom 𝐹 = B))
5	2, 3, 4	3bitr4g 212	1 ⊢ (A = B → (𝐹 Fn A ↔ 𝐹 Fn B))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242  dom cdm 4288  Fun wfun 4839   Fn wfn 4840

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-4 1397  ax-17 1416  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-cleq 2030  df-fn 4848

This theorem is referenced by:  fneq2d  4933  fneq2i  4937  feq2  4974  foeq2  5046  f1o00  5104  eqfnfv2  5209  tfr0  5878  tfrlemisucaccv  5880  tfrlemi1  5887  tfrlemi14d  5888  tfrexlem  5889  0fz1  8679