
theorem Th67:
  for C being Coherence_Space, x,y being set st x <> y & {x,y} in
  C holds not {x,y} in 'not' C
proof
  let C be Coherence_Space, x,y be set;
  assume that
A1: x <> y and
A2: {x,y} in C & {x,y} in 'not' C;
  consider z being set such that
A3: {x,y} /\ {x,y} c= {z} by A2,Th65;
  x = z by A3,ZFMISC_1:20;
  hence contradiction by A1,A3,ZFMISC_1:20;
end;
