
theorem
for G being void SimpleGraph holds clique# G = 0
proof
 let G be void SimpleGraph;
 assume A1: clique# G <> 0;
   consider C being finite Clique of G such that
A2: order C = clique# G by Def15;
   Vertices C c= Vertices G by ZFMISC_1:77;
   hence contradiction by A1,A2;
end;
