Theorem feq3 2750

Description: Equality theorem for functions.

Assertion

Ref	Expression
feq3	|- (A = B -> (F:C-->A <-> F:C-->B))

Proof of Theorem feq3

Step	Hyp	Ref	Expression
1		sseq2 1522	. . 3 |- (A = B -> (ran F (_ A <-> ran F (_ B))
2	1	anbi2d 468	. 2 |- (A = B -> ((F Fn C /\ ran F (_ A) <-> (F Fn C /\ ran F (_ B)))
3		df-f 2434	. 2 |- (F:C-->A <-> (F Fn C /\ ran F (_ A))
4		df-f 2434	. 2 |- (F:C-->B <-> (F Fn C /\ ran F (_ B))
5	2, 3, 4	3bitr4g 428	1 |- (A = B -> (F:C-->A <-> F:C-->B))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196   = wceq 1091   (_ wss 1487  ran crn 2411   Fn wfn 2417  -->wf 2418

This theorem is referenced by:  fconstg 2775  f1eq3 2778  fsn2 2896  mapvalg 3263  mapsn 3269  mapdom2 3389  mapunen 3397  closedsub 5128  ch2 5149

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-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492  df-f 2434

metamath.org