Theorem disjel 3458

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 3456	. . 3 |- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2		eleq2 2228	. . . 4 |- ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3		eldifn 3240	. . . 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 1342   e. wcel 2135   \ cdif 3108   i^i cin 3110   (/)c0 3404

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 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-bndl 1496  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146

This theorem depends on definitions:  df-bi 116  df-tru 1345  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-nfc 2295  df-ral 2447  df-v 2723  df-dif 3113  df-in 3117  df-nul 3405

This theorem is referenced by:  fvun1  5546  ctssdccl  7067  fsumsplit  11334  fprodsplitdc  11523  fprodsplit  11524