Theorem fneq2 5174

Description: Equality theorem for function predicate with domain. (Contributed by set.mm contributors, 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 2362	. . 3 ⊢ (A = B → (dom F = A ↔ dom F = B))
2	1	anbi2d 684	. 2 ⊢ (A = B → ((Fun F ∧ dom F = A) ↔ (Fun F ∧ dom F = B)))
3		df-fn 4790	. 2 ⊢ (F Fn A ↔ (Fun F ∧ dom F = A))
4		df-fn 4790	. 2 ⊢ (F Fn B ↔ (Fun F ∧ dom F = B))
5	2, 3, 4	3bitr4g 279	1 ⊢ (A = B → (F Fn A ↔ F Fn B))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642  dom cdm 4772  Fun wfun 4775   Fn wfn 4776

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-cleq 2346  df-fn 4790

This theorem is referenced by:  fneq2d  5176  fneq2i  5179  feq2  5211  foeq2  5266  f1o00  5317  eqfnfv2  5393  fconstfv  5456  composefn  5818  brfns  5833  fnfullfun  5858