>>

>>	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.

Yes, and start reading the question first