Theorem intex 2729

Description: The intersection of a non-empty class exists. Exercise 5 of [TakeutiZaring] p. 44 and its converse.

Assertion

Ref	Expression
intex	|- (A =/= (/) <-> |^|A e. V)

Proof of Theorem intex

Step	Hyp	Ref	Expression
1		ne0 2288	. . 3 |- (A =/= (/) <-> E.x x e. A)
2		intss1 2548	. . . . 5 |- (x e. A -> |^|A (_ x)
3		visset 1813	. . . . . 6 |- x e. V
4	3	ssex 2719	. . . . 5 |- (|^|A (_ x -> |^|A e. V)
5	2, 4	syl 10	. . . 4 |- (x e. A -> |^|A e. V)
6	5	19.23aiv 1295	. . 3 |- (E.x x e. A -> |^|A e. V)
7	1, 6	sylbi 199	. 2 |- (A =/= (/) -> |^|A e. V)
8		nvelv 2713	. . . 4 |- -. V e. V
9		inteq 2536	. . . . . 6 |- (A = (/) -> |^|A = |^|(/))
10		int0 2547	. . . . . 6 |- |^|(/) = V
11	9, 10	syl6eq 1523	. . . . 5 |- (A = (/) -> |^|A = V)
12	11	eleq1d 1540	. . . 4 |- (A = (/) -> (|^|A e. V <-> V e. V))
13	8, 12	mtbiri 717	. . 3 |- (A = (/) -> -. |^|A e. V)
14	13	necon2ai 1611	. 2 |- (|^|A e. V -> A =/= (/))
15	7, 14	impbi 157	1 |- (A =/= (/) <-> |^|A e. V)

Syntax hints:   <-> wb 146   = wceq 956   e. wcel 958  E.wex 980   =/= wne 1585  Vcvv 1811   (_ wss 2047  (/)c0 2280  |^|cint 2533

This theorem is referenced by:  intnex 2730  intexab 2731  onint0 3007  onintrab 3013  onmindif2 3061  abfii2OLD 4562  tz9.12lem1 4659  tz9.12lem3 4661  rankval 4668  oncardon 4820  oncardid 4821  cardon 4827  cardid 4828  cardcf 4911  subbas2OLD 7645  hsupval2t 9300  hsupclt 9307  fiv 10482  fivOLD 10483

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-13 969  ax-14 970  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459  ax-sep 2703

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-ral 1649  df-v 1812  df-dif 2049  df-in 2051  df-ss 2053  df-nul 2281  df-int 2534

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