theorem Th26:
  1 in a & 1 in b implies a in a |^|^ b proof assume
A1: 1 in a;
    assume 1 in b; then
A2: succ 1 c= b by ORDINAL1:21;
    0 in Segm 1 by NAT_1:44; then
 0 in a by A1,ORDINAL1:10;
    then a|^|^1 in a|^|^2 & a|^|^2 c= a|^|^b by A1,A2,Th21,Th24; then
    a|^|^1 in a|^|^b;
    hence a in a |^|^ b by Th16;
  end;
