Theorem fneq2 5348

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 2206	. . 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 5262	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 5262	. 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 1364  dom cdm 4664  Fun wfun 5253   Fn wfn 5254

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 1461  ax-gen 1463  ax-4 1524  ax-17 1540  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-cleq 2189  df-fn 5262

This theorem is referenced by:  fneq2d  5350  fneq2i  5354  feq2  5392  foeq2  5478  f1o00  5540  eqfnfv2  5661  tfr0dm  6381  tfrlemisucaccv  6384  tfrlemi1  6391  tfrlemi14d  6392  tfrexlem  6393  tfr1onlemsucfn  6399  tfr1onlemsucaccv  6400  tfr1onlembxssdm  6402  tfr1onlembfn  6403  tfr1onlemaccex  6407  tfr1onlemres  6408  ixpeq1  6769  0fz1  10122