Theorem fn0 5503

Description: A function with empty domain is empty. (Contributed by NM, 15-Apr-1998.) (Proof shortened by Andrew Salmon, 17-Sep-2011.)

Assertion

Ref	Expression
fn0	|- ( F Fn (/) <-> F = (/) )

Proof of Theorem fn0

Step	Hyp	Ref	Expression
1		fnrel 5479	. . 3 |- ( F Fn (/) -> Rel F )
2		fndm 5480	. . 3 |- ( F Fn (/) -> dom F = (/) )
3		reldm0 4999	. . . 4 |- ( Rel F -> ( F = (/) <-> dom F = (/) ) )
4	3	biimpar 297	. . 3 |- ( ( Rel F /\ dom F = (/) ) -> F = (/) )
5	1, 2, 4	syl2anc 415	. 2 |- ( F Fn (/) -> F = (/) )
6		fun0 5439	. . . 4 |- Fun (/)
7		dm0 4995	. . . 4 |- dom (/) = (/)
8		df-fn 5380	. . . 4 |- ( (/) Fn (/) <-> ( Fun (/) /\ dom (/) = (/) ) )
9	6, 7, 8	mpbir2an 955	. . 3 |- (/) Fn (/)
10		fneq1 5469	. . 3 |- ( F = (/) -> ( F Fn (/) <-> (/) Fn (/) ) )
11	9, 10	mpbiri 168	. 2 |- ( F = (/) -> F Fn (/) )
12	5, 11	impbii 126	1 |- ( F Fn (/) <-> F = (/) )

This proof depends on syntax axioms:   <-> wb 105   = wceq 1402   (/)c0 3520   dom cdm 4774   Rel wrel 4779   Fun wfun 5371   Fn wfn 5372

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-nul 4259  ax-pow 4311  ax-pr 4346

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-nul 3521  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-br 4131  df-opab 4193  df-id 4438  df-xp 4780  df-rel 4781  df-cnv 4782  df-co 4783  df-dm 4784  df-fun 5379  df-fn 5380

This theorem is used by:  mpt0  5511  f0  5583  f00  5584  f0bi  5585  f1o00  5676  fo00  5677  tpos0  6545  ixp0x  7008  0fz1  10449  hashf1  11287