Theorem disjel 3298

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 3296	. . 3 |- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2		eleq2 2142	. . . 4 |- ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3		eldifn 3095	. . . 4 |- ( C e. ( A \ B ) -> -. C e. B )
4	2, 3	syl6bi 161	. . 3 |- ( A = ( A \ B ) -> ( C e. A -> -. C e. B ) )
5	1, 4	sylbi 119	. 2 |- ( ( A i^i B ) = (/) -> ( C e. A -> -. C e. B ) )
6	5	imp 122	1 |- ( ( ( A i^i B ) = (/) /\ C e. A ) -> -. C e. B )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 102   = wceq 1284   e. wcel 1433   \ cdif 2970   i^i cin 2972   (/)c0 3251

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-v 2603  df-dif 2975  df-in 2979  df-nul 3252

This theorem is referenced by:  fvun1  5260