Theorem disj4 3599

Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)

Assertion

Ref	Expression
disj4	|- ((A i^i B) = (/) <-> -. (A \ B) C. A)

Proof of Theorem disj4

Step	Hyp	Ref	Expression
1		disj3 3595	. 2 |- ((A i^i B) = (/) <-> A = (A \ B))
2		eqcom 2355	. 2 |- (A = (A \ B) <-> (A \ B) = A)
3		difss 3393	. . . 4 |- (A \ B) C_ A
4		dfpss2 3354	. . . 4 |- ((A \ B) C. A <-> ((A \ B) C_ A /\ -. (A \ B) = A))
5	3, 4	mpbiran 884	. . 3 |- ((A \ B) C. A <-> -. (A \ B) = A)
6	5	con2bii 322	. 2 |- ((A \ B) = A <-> -. (A \ B) C. A)
7	1, 2, 6	3bitri 262	1 |- ((A i^i B) = (/) <-> -. (A \ B) C. A)

Syntax hints:  -. wn 3   <-> wb 176   = wceq 1642   \ cdif 3206   i^i cin 3208   C_ wss 3257   C. wpss 3258  (/)c0 3550

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 2478  df-ne 2518  df-ral 2619  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259  df-pss 3261  df-nul 3551

This theorem is referenced by: (None)