Theorem disjel 3505

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 3503	. . 3 |- ( ( A i^i B ) = (/) <-> A = ( A \ B ) )
2		eleq2 2260	. . . 4 |- ( A = ( A \ B ) -> ( C e. A <-> C e. ( A \ B ) ) )
3		eldifn 3286	. . . 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 1364   e. wcel 2167   \ cdif 3154   i^i cin 3156   (/)c0 3450

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 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-v 2765  df-dif 3159  df-in 3163  df-nul 3451

This theorem is referenced by:  fvun1  5627  ctssdccl  7177  fsumsplit  11572  fprodsplitdc  11761  fprodsplit  11762