
theorem Th118:
for n being Nat
 holds clique# ((MycielskianSeq CompleteSGraph 2).n) = 2 &
       chromatic# ((MycielskianSeq CompleteSGraph 2).n) = n+2
proof
A1: card 2 = 2;
  set P2 = CompleteSGraph 2;
  defpred P[Nat] means
   clique# ((MycielskianSeq P2).$1) = 2 &
   chromatic# ((MycielskianSeq P2).$1) = $1+2;
A2: clique# ((MycielskianSeq P2).0) = clique# P2 by Th113 .= 2 by A1,Th59;
   chromatic# ((MycielskianSeq P2).0) = chromatic# P2 by Th113
    .= 0+2 by A1,Th69;
   then
A3: P[0] by A2;
A4: for n being Nat st P[n] holds P[n+1] proof
     let n be Nat;
     assume A5: P[n];
      thus clique# ((MycielskianSeq P2).(n+1))
         = clique# Mycielskian((MycielskianSeq P2).n) by Th115
        .= 2 by Th110,A5;
      thus chromatic# ((MycielskianSeq P2).(n+1))
         = chromatic# Mycielskian((MycielskianSeq P2).n) by Th115
        .= 1+(n+2) by A5,Th112
        .= (n+1)+2;
    end;
 for n being Nat holds P[n] from NAT_1:sch 2(A3,A4);
 hence thesis;
end;
