Theorem ssdifin0 3632

Description: A subset of a difference does not intersect the subtrahend. (Contributed by Jeff Hankins, 1-Sep-2013.) (Proof shortened by Mario Carneiro, 24-Aug-2015.)

Assertion

Ref	Expression
ssdifin0	|- (A C_ (B \ C) -> (A i^i C) = (/))

Proof of Theorem ssdifin0

Step	Hyp	Ref	Expression
1		ssrin 3481	. 2 |- (A C_ (B \ C) -> (A i^i C) C_ ((B \ C) i^i C))
2		incom 3449	. . 3 |- ((B \ C) i^i C) = (C i^i (B \ C))
3		disjdif 3623	. . 3 |- (C i^i (B \ C)) = (/)
4	2, 3	eqtri 2373	. 2 |- ((B \ C) i^i C) = (/)
5		sseq0 3583	. 2 |- (((A i^i C) C_ ((B \ C) i^i C) /\ ((B \ C) i^i C) = (/)) -> (A i^i C) = (/))
6	1, 4, 5	sylancl 643	1 |- (A C_ (B \ C) -> (A i^i C) = (/))

Syntax hints:   -> wi 4   = wceq 1642   \ 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-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:  ssdifeq0  3633