theorem Th69:
  p in s & s in q implies (T,p,q)incl.(p,s) = [p,s] & (T,p,q)incl.(s,q) = [s,q]
  proof assume
    p in s & s in q; then
    p c= s & s c= q by ORDINAL1:def 2;
    hence thesis by Th62;
  end;
