Theorem disjel 3598

Description: A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)

Assertion

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

Proof of Theorem disjel

Step	Hyp	Ref	Expression
1		disj3 3596	. . 3 |- ((A i^i B) = (/) <-> A = (A \ B))
2		eleq2 2414	. . . 4 |- (A = (A \ B) -> (C e. A <-> C e. (A \ B)))
3		eldifn 3390	. . . 4 |- (C e. (A \ B) -> -. C e. B)
4	2, 3	syl6bi 219	. . 3 |- (A = (A \ B) -> (C e. A -> -. C e. B))
5	1, 4	sylbi 187	. 2 |- ((A i^i B) = (/) -> (C e. A -> -. C e. B))
6	5	imp 418	1 |- (((A i^i B) = (/) /\ C e. A) -> -. C e. B)

Syntax hints:  -. wn 3   -> wi 4   /\ wa 358   = wceq 1642   e. wcel 1710   \ cdif 3207   i^i cin 3209  (/)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-nul 3552

This theorem is referenced by: (None)