theorem Th61:
  subdivision(0,P,KX) = KX
 proof
  ex F be Function st F.0=KX & F.0=subdivision(0,P,KX) & dom F=NAT & for k for
KX1 be SimplicialComplexStr of X st KX1=F.k holds
F.(k+1)=subdivision(P,KX1) by Def21;
  hence thesis;
 end;
