>

>	select S.* ;
>		from Schnaps S, ;
>			( ;
>				select c3 ;
>					from Schnaps ;
>					group by  1 ;
>					having ( count( distinct c1 + c2 ) > 1) ;
>			) AA ;
>			where	( S.c3 == AA.c3 )
>		
>

>
>

Unfortunately, this would not give us only error cases - I just thought of the same SQL. You don't need to join here. This would give all C3 cases even when a/b combination is the same - I don't see a simple way to separate them - need to think.

UPDATE. Actually, it may be simple - we just need to add min(a) and min(b).