Theorem disj 2311

Description: Two ways of saying that two classes are disjoint (have no members in common).

Assertion

Ref	Expression
disj	|- ((A i^i B) = (/) <-> A.x e. A -. x e. B)

Distinct variable groups:   x,A   x,B

Proof of Theorem disj

Step	Hyp	Ref	Expression
1		df-in 2051	. . . 4 |- (A i^i B) = {x | (x e. A /\ x e. B)}
2	1	eqeq1i 1482	. . 3 |- ((A i^i B) = (/) <-> {x | (x e. A /\ x e. B)} = (/))
3		abeq1 1569	. . 3 |- ({x | (x e. A /\ x e. B)} = (/) <-> A.x((x e. A /\ x e. B) <-> x e. (/)))
4		imnan 242	. . . . 5 |- ((x e. A -> -. x e. B) <-> -. (x e. A /\ x e. B))
5		noel 2284	. . . . . 6 |- -. x e. (/)
6	5	nbn 722	. . . . 5 |- (-. (x e. A /\ x e. B) <-> ((x e. A /\ x e. B) <-> x e. (/)))
7	4, 6	bitr2 174	. . . 4 |- (((x e. A /\ x e. B) <-> x e. (/)) <-> (x e. A -> -. x e. B))
8	7	albii 999	. . 3 |- (A.x((x e. A /\ x e. B) <-> x e. (/)) <-> A.x(x e. A -> -. x e. B))
9	2, 3, 8	3bitr 177	. 2 |- ((A i^i B) = (/) <-> A.x(x e. A -> -. x e. B))
10		df-ral 1649	. 2 |- (A.x e. A -. x e. B <-> A.x(x e. A -> -. x e. B))
11	9, 10	bitr4 176	1 |- ((A i^i B) = (/) <-> A.x e. A -. x e. B)

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223  A.wal 954   = wceq 956   e. wcel 958  {cab 1463  A.wral 1645   i^i cin 2046  (/)c0 2280

This theorem is referenced by:  disj1 2312  disjne 2315  dffr2 2919  onint 3006  onxpdisj 3241  zfreg 4596  zfreg2 4597  kmlem4 4768  ssxr 5540  qdensere 7751  bl2in 7843  lpbl 7880

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-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

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-ral 1649  df-v 1812  df-dif 2049  df-in 2051  df-nul 2281

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