Theorem fneq2 6280

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 2789	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 619	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6193	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6193	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 306	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 387   = wceq 1507  dom cdm 5408  Fun wfun 6184   Fn wfn 6185

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-9 2059  ax-ext 2750

This theorem depends on definitions:  df-bi 199  df-an 388  df-ex 1743  df-cleq 2771  df-fn 6193

This theorem is referenced by:  fneq2d  6282  fneq2i  6286  feq2  6328  foeq2  6418  f1o00  6480  eqfnfv2  6630  wfrlem1  7759  wfrlem15  7775  tfrlem12  7831  ixpeq1  8272  ac5  9699  0fz1  12746  esumcvgsum  30991  bnj90  31640  bnj919  31686  bnj535  31809  bnj1463  31972  frrlem1  32644  frrlem13  32656  fnchoice  40705