Theorem fneq2 2719

Description: Equality theorem for function predicate with domain.

Assertion

Ref	Expression
fneq2	|- (A = B -> (F Fn A <-> F Fn B))

Proof of Theorem fneq2

Step	Hyp	Ref	Expression
1		cleq2 1110	. . 3 |- (A = B -> (dom F = A <-> dom F = B))
2	1	anbi2d 468	. 2 |- (A = B -> ((Fun F /\ dom F = A) <-> (Fun F /\ dom F = B)))
3		df-fn 2433	. 2 |- (F Fn A <-> (Fun F /\ dom F = A))
4		df-fn 2433	. 2 |- (F Fn B <-> (Fun F /\ dom F = B))
5	2, 3, 4	3bitr4g 428	1 |- (A = B -> (F Fn A <-> F Fn B))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196   = wceq 1091  dom cdm 2410  Fun wfun 2416   Fn wfn 2417

This theorem is referenced by:  feq2 2749  foeq2 2785  f1o00 2823  cleqfv 2880  fconstfv 2903  tfrlem3 2951  tfrlem12 2960  fnoprab2 3039  aceq3 3556  ac7g 3570  ac5 3573  fodom 3613

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-cleq 1097  df-fn 2433

metamath.org