Theorem feq2 5212

Description: Equality theorem for functions. (Contributed by set.mm contributors, 1-Aug-1994.)

Assertion

Ref	Expression
feq2	|- (A = B -> (F:A-->C <-> F:B-->C))

Proof of Theorem feq2

Step	Hyp	Ref	Expression
1		fneq2 5175	. . 3 |- (A = B -> (F Fn A <-> F Fn B))
2	1	anbi1d 685	. 2 |- (A = B -> ((F Fn A /\ ran F C_ C) <-> (F Fn B /\ ran F C_ C)))
3		df-f 4792	. 2 |- (F:A-->C <-> (F Fn A /\ ran F C_ C))
4		df-f 4792	. 2 |- (F:B-->C <-> (F Fn B /\ ran F C_ C))
5	2, 3, 4	3bitr4g 279	1 |- (A = B -> (F:A-->C <-> F:B-->C))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358   = wceq 1642   C_ wss 3258  ran crn 4774   Fn wfn 4777  -->wf 4778

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 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 4791  df-f 4792

This theorem is referenced by:  feq23  5214  feq2d  5216  feq2i  5219  f00  5250  f1eq2  5255  fressnfv  5440  fconstfv  5457  mapex  6007  mapvalg  6010