Theorem reldisj 2313

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 2063	. . . . 5 |- (A (_ C -> (x e. A -> x e. C))
2	1	biantrurd 727	. . . 4 |- (A (_ C -> ((x e. A -> -. x e. B) <-> ((x e. A -> x e. C) /\ (x e. A -> -. x e. B))))
3		pm4.76 599	. . . . 5 |- (((x e. A -> x e. C) /\ (x e. A -> -. x e. B)) <-> (x e. A -> (x e. C /\ -. x e. B)))
4		eldif 2057	. . . . . 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 536	. . 3 |- (A (_ C -> ((x e. A -> -. x e. B) <-> (x e. A -> x e. (C \ B))))
8	7	albidv 1278	. 2 |- (A (_ C -> (A.x(x e. A -> -. x e. B) <-> A.x(x e. A -> x e. (C \ B))))
9		disj1 2312	. 2 |- ((A i^i B) = (/) <-> A.x(x e. A -> -. x e. B))
10		dfss2 2058	. 2 |- (A (_ (C \ B) <-> A.x(x e. A -> x e. (C \ B)))
11	8, 9, 10	3bitr4g 555	1 |- (A (_ C -> ((A i^i B) = (/) <-> A (_ (C \ B)))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223  A.wal 954   = wceq 956   e. wcel 958   \ cdif 2044   i^i cin 2046   (_ wss 2047  (/)c0 2280

This theorem is referenced by:  disj2 2316  elcls 7704  islp2 7747

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-ss 2053  df-nul 2281

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