Theorem reldisj 2309

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C.

Assertion

Ref	Expression
reldisj	|- (A (_ C -> ((A i^i B) = (/) <-> A (_ (C \ B)))

Proof of Theorem reldisj

Step	Hyp	Ref	Expression
1		ssel 2059	. . . . 5 |- (A (_ C -> (x e. A -> x e. C))
2	1	biantrurd 726	. . . 4 |- (A (_ C -> ((x e. A -> -. x e. B) <-> ((x e. A -> x e. C) /\ (x e. A -> -. x e. B))))
3		pm4.76 598	. . . . 5 |- (((x e. A -> x e. C) /\ (x e. A -> -. x e. B)) <-> (x e. A -> (x e. C /\ -. x e. B)))
4		eldif 2053	. . . . . 6 |- (x e. (C \ B) <-> (x e. C /\ -. x e. B))
5	4	imbi2i 185	. . . . 5 |- ((x e. A -> x e. (C \ B)) <-> (x e. A -> (x e. C /\ -. x e. B)))
6	3, 5	bitr4 176	. . . 4 |- (((x e. A -> x e. C) /\ (x e. A -> -. x e. B)) <-> (x e. A -> x e. (C \ B)))
7	2, 6	syl6bb 535	. . 3 |- (A (_ C -> ((x e. A -> -. x e. B) <-> (x e. A -> x e. (C \ B))))
8	7	albidv 1276	. 2 |- (A (_ C -> (A.x(x e. A -> -. x e. B) <-> A.x(x e. A -> x e. (C \ B))))
9		disj1 2308	. 2 |- ((A i^i B) = (/) <-> A.x(x e. A -> -. x e. B))
10		dfss2 2054	. 2 |- (A (_ (C \ B) <-> A.x(x e. A -> x e. (C \ B)))
11	8, 9, 10	3bitr4g 554	1 |- (A (_ C -> ((A i^i B) = (/) <-> A (_ (C \ B)))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223  A.wal 952   = wceq 954   e. wcel 956   \ cdif 2040   i^i cin 2042   (_ wss 2043  (/)c0 2276

This theorem is referenced by:  disj2 2312  elcls 7654  islp2 7697

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-ral 1646  df-v 1808  df-dif 2045  df-in 2047  df-ss 2049  df-nul 2277

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