Theorem disjel 3463

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 3461	. . 3 |- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2		eleq2 2230	. . . 4 |- ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3		eldifn 3245	. . . 4 |- ( C e. ( A \ B ) -> -. C e. B )
4	2, 3	syl6bi 162	. . 3 |- ( A = ( A \ B ) -> ( C e. A -> -. C e. B ) )
5	1, 4	sylbi 120	. 2 |- ( ( A i^i B ) = (/) -> ( C e. A -> -. C e. B ) )
6	5	imp 123	1 |- ( ( ( A i^i B ) = (/) /\ C e. A ) -> -. C e. B )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 103   = wceq 1343   e. wcel 2136   \ cdif 3113   i^i cin 3115   (/)c0 3409

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-v 2728  df-dif 3118  df-in 3122  df-nul 3410

This theorem is referenced by:  fvun1  5552  ctssdccl  7076  fsumsplit  11348  fprodsplitdc  11537  fprodsplit  11538