Theorem intex 2725

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 2285	. . 3 |- (A =/= (/) <-> E.x x e. A)
2		intss1 2544	. . . . 5 |- (x e. A -> |^|A (_ x)
3		visset 1810	. . . . . 6 |- x e. V
4	3	ssex 2715	. . . . 5 |- (|^|A (_ x -> |^|A e. V)
5	2, 4	syl 10	. . . 4 |- (x e. A -> |^|A e. V)
6	5	19.23aiv 1294	. . 3 |- (E.x x e. A -> |^|A e. V)
7	1, 6	sylbi 199	. 2 |- (A =/= (/) -> |^|A e. V)
8		nvelv 2709	. . . 4 |- -. V e. V
9		inteq 2532	. . . . . 6 |- (A = (/) -> |^|A = |^|(/))
10		int0 2543	. . . . . 6 |- |^|(/) = V
11	9, 10	syl6eq 1521	. . . . 5 |- (A = (/) -> |^|A = V)
12	11	eleq1d 1538	. . . 4 |- (A = (/) -> (|^|A e. V <-> V e. V))
13	8, 12	mtbiri 716	. . 3 |- (A = (/) -> -. |^|A e. V)
14	13	necon2ai 1609	. 2 |- (|^|A e. V -> A =/= (/))
15	7, 14	impbi 157	1 |- (A =/= (/) <-> |^|A e. V)

Syntax hints:   <-> wb 146   = wceq 955   e. wcel 957  E.wex 979   =/= wne 1583  Vcvv 1808   (_ wss 2044  (/)c0 2277  |^|cint 2529

This theorem is referenced by:  intnex 2726  intexab 2727  onint0 3003  onintrab 3009  onmindif2 3057  abfii2 4545  tz9.12lem1 4642  tz9.12lem3 4644  rankval 4651  oncardon 4803  oncardid 4804  cardon 4810  cardid 4811  cardcf 4894  subbas2 7605  hsupval2t 9255  hsupclt 9262  fiv 10433

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-13 968  ax-14 969  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217  ax-ext 1458  ax-sep 2699

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 980  df-sb 1171  df-clab 1463  df-cleq 1468  df-clel 1471  df-ne 1585  df-ral 1647  df-v 1809  df-dif 2046  df-in 2048  df-ss 2050  df-nul 2278  df-int 2530

Copyright terms: Public domain