>>
>> 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
Gregory