theorem Th15:
  not a,b // c,d & a,b // p,q & c,d // r,s & p<>q & r<>s implies not p,q // r,s
proof
  assume that
A1: not a,b // c,d and
A2: a,b // p,q and
A3: c,d // r,s and
A4: p<>q and
A5: r<>s;
  assume p,q // r,s;
  then a,b // r,s by A2,A4,Th8;
  then
A6: r,s // a,b by Th6;
  r,s // c,d by A3,Th6;
  hence contradiction by A1,A5,A6,Def1;
end;
