Theorem fneq2 6639

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 2745	. . 3 ⊢ (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
2	1	anbi2d 630	. 2 ⊢ (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3		df-fn 6544	. 2 ⊢ (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4		df-fn 6544	. 2 ⊢ (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
5	2, 3, 4	3bitr4g 314	1 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 397   = wceq 1542  dom cdm 5676  Fun wfun 6535   Fn wfn 6536

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-ex 1783  df-cleq 2725  df-fn 6544

This theorem is referenced by:  fneq2d  6641  fneq2i  6645  feq2  6697  foeq2  6800  f1o00  6866  eqfnfv2  7031  frrlem1  8268  frrlem13  8280  wfrlem1OLD  8305  wfrlem15OLD  8320  tfrlem12  8386  ixpeq1  8899  ac5  10469  0fz1  13518  esumcvgsum  33075  bnj90  33722  bnj919  33767  bnj535  33890  bnj1463  34055  fnchoice  43699