Theorem disjel 3578

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 3576	. . 3 |- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2		eleq2 2302	. . . 4 |- ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3		eldifn 3352	. . . 4 |- ( C e. ( A \ B ) -> -. C e. B )
4	2, 3	biimtrdi 163	. . 3 |- ( A = ( A \ B ) -> ( C e. A -> -. C e. B ) )
5	1, 4	sylbi 121	. 2 |- ( ( A i^i B ) = (/) -> ( C e. A -> -. C e. B ) )
6	5	imp 124	1 |- ( ( ( A i^i B ) = (/) /\ C e. A ) -> -. C e. B )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104   = wceq 1402   e. wcel 2209   \ cdif 3217   i^i cin 3219   (/)c0 3520

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-v 2823  df-dif 3222  df-in 3226  df-nul 3521

This theorem is referenced by:  fvun1  5763  ctssdccl  7441  fsumsplit  12152  fprodsplitdc  12341  fprodsplit  12342