
theorem Th68: :: Simplicial03a
  for G being _Graph, v being Vertex of G st not v is simplicial
  ex a,b being Vertex of G st a<>b & v<>a & v<>b & v,a are_adjacent & v,b
  are_adjacent & not a,b are_adjacent
proof
  let G be _Graph, v be Vertex of G such that
A1: not v is simplicial;
  assume
A2: not ex a,b being Vertex of G st a<>b & v<>a & v<>b & v,a
  are_adjacent & v,b are_adjacent & not a,b are_adjacent;
  per cases;
  suppose
    G.AdjacentSet({v}) = {};
    hence contradiction by A1;
  end;
  suppose
A3: G.AdjacentSet({v}) <> {};
    now
      let H be AdjGraph of G,{v};
A4:   H is inducedSubgraph of G,G.AdjacentSet({v}) by Def5;
      now
        let a,b be Vertex of H such that
A5:     a<>b;
A6:     the_Vertices_of H = G.AdjacentSet({v}) by A3,A4,GLIB_000:def 37;
        then
A7:     b in G.AdjacentSet({v});
        a in G.AdjacentSet({v}) by A6;
        then reconsider vv=v,aa=a,bb=b as Vertex of G by A7;
A8:     aa,vv are_adjacent by A6,Th51;
A9:     bb,vv are_adjacent by A6,Th51;
        aa<>vv by A6,Th51;
        then aa,bb are_adjacent by A2,A5,A8,A9;
        hence a,b are_adjacent by A3,A4,Th44;
      end;
      hence H is complete;
    end;
    hence contradiction by A1;
  end;
end;
