Theorem reldisj 3595

Description: Two ways of saying that two classes are disjoint, using the complement of B relative to a universe C. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem reldisj

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfss2 3263	. . . 4 |- (A C_ C <-> A.x(x e. A -> x e. C))
2		pm5.44 877	. . . . . 6 |- ((x e. A -> x e. C) -> ((x e. A -> -. x e. B) <-> (x e. A -> (x e. C /\ -. x e. B))))
3		eldif 3222	. . . . . . 7 |- (x e. (C \ B) <-> (x e. C /\ -. x e. B))
4	3	imbi2i 303	. . . . . 6 |- ((x e. A -> x e. (C \ B)) <-> (x e. A -> (x e. C /\ -. x e. B)))
5	2, 4	syl6bbr 254	. . . . 5 |- ((x e. A -> x e. C) -> ((x e. A -> -. x e. B) <-> (x e. A -> x e. (C \ B))))
6	5	sps 1754	. . . 4 |- (A.x(x e. A -> x e. C) -> ((x e. A -> -. x e. B) <-> (x e. A -> x e. (C \ B))))
7	1, 6	sylbi 187	. . 3 |- (A C_ C -> ((x e. A -> -. x e. B) <-> (x e. A -> x e. (C \ B))))
8	7	albidv 1625	. 2 |- (A C_ C -> (A.x(x e. A -> -. x e. B) <-> A.x(x e. A -> x e. (C \ B))))
9		disj1 3594	. 2 |- ((A i^i B) = (/) <-> A.x(x e. A -> -. x e. B))
10		dfss2 3263	. 2 |- (A C_ (C \ B) <-> A.x(x e. A -> x e. (C \ B)))
11	8, 9, 10	3bitr4g 279	1 |- (A C_ C -> ((A i^i B) = (/) <-> A C_ (C \ B)))

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176   /\ wa 358  A.wal 1540   = wceq 1642   e. wcel 1710   \ cdif 3207   i^i cin 3209   C_ wss 3258  (/)c0 3551

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by:  disj2  3599