
theorem Th104:
for G being SimpleGraph holds G = (Mycielskian G) SubgraphInducedBy Vertices G
proof
 let G be SimpleGraph;
 set L = Vertices G, MG = Mycielskian G;
 thus G c= MG SubgraphInducedBy L by Th84,Th44;
 thus MG SubgraphInducedBy L c= G proof
   let a be object;
   assume A1: a in MG SubgraphInducedBy L;
   reconsider m = a as set by TARSKI:1;
   A2: m in bool L by A1,XBOOLE_0:def 4;
   per cases by A1,MYCIELSK:4;
   suppose m in { {} };
      then m = {} by TARSKI:def 1;
    hence a in G by Th20;
   end;
   suppose m in the set of all {x} where x is Element of L \/ [:L,{L}:] \/ {L};
      then consider x being Element of L \/ [:L,{L}:] \/ {L} such that
   A3: m = {x};
       x in m by A3,TARSKI:def 1;
    hence a in G by A2,A3,Th24;
   end;
   suppose m in Edges G;
    hence a in G;
   end;
   suppose  m in { {x,[y,L]} where x, y is Element of L :
            {x,y} in Edges G };
     then consider x, y being Element of L such that
   A4: m = {x,[y,L]} and {x,y} in Edges G;
       [y,L] in m by A4,TARSKI:def 2;
    hence a in G by A2,Th1;
   end;
   suppose m in { {L,[x,L]} where x is Element of L : x in Vertices G };
     then consider x being Element of L such that
   A5: m = {L,[x,L]} and x in Vertices G;
       L in m by A5,TARSKI:def 2;
       then L in L by A2;
    hence a in G;
   end;
 end;
end;
