Theorem disj 2315

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 2054	. . . 4 |- (A i^i B) = {x | (x e. A /\ x e. B)}
2	1	eqeq1i 1485	. . 3 |- ((A i^i B) = (/) <-> {x | (x e. A /\ x e. B)} = (/))
3		abeq1 1572	. . 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 2287	. . . . . 6 |- -. x e. (/)
6	5	nbn 724	. . . . 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 1001	. . 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 1652	. 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 956   = wceq 958   e. wcel 960  {cab 1466  A.wral 1648   i^i cin 2049  (/)c0 2283

This theorem is referenced by:  disj1 2316  disjne 2319  dffr2 2925  onint 3012  onxpdisj 3247  zfreg 4605  zfreg2 4606  kmlem4 4778  ssxr 5552  qdensere 7748  bl2in 7840  lpbl 7877

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ral 1652  df-v 1815  df-dif 2052  df-in 2054  df-nul 2284

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