reserve
  a,b,c,d,e for Ordinal,
  m,n for Nat,
  f for Ordinal-Sequence,
  x for object;
reserve S,S1,S2 for Sequence;

theorem Th24:
  1 in a & m < n implies a |^|^ m in a |^|^ n proof assume
A1: 1 in a & m < n; then m+1 <= n by NAT_1:13; then
    consider k being Nat such that
A2: n = m+1+k by NAT_1:10;
    defpred Q[Nat] means a|^|^$1 in a|^|^($1+1);
    defpred P[Nat] means a |^|^ m in a |^|^ (m+1+$1);
    a|^|^0 = 1 by Th13; then
A3: Q[0] by A1,Th16;
A4: now let n; assume
A5:   Q[n];
      succ Segm n = Segm(n+1) & succ Segm (n+1) = Segm(n+1+1) by NAT_1:38;
then
      a|^|^(n+1) = exp(a, a|^|^n) & a|^|^((n+1)+1) = exp(a, a|^|^(n+1)) by Th14
;
      hence Q[n+1] by A5,A1,ORDINAL4:24;
    end;
A6: for n holds Q[n] from NAT_1:sch 2(A3,A4); then
A7: P[0];
A8: now let k be Nat; assume
A9:   P[k];
      a|^|^(m+1+k) in a|^|^((m+1+k)+1) by A6;
      hence P[k+1] by A9,ORDINAL1:10;
    end;
    for k being Nat holds P[k] from NAT_1:sch 2(A7,A8);
    hence thesis by A2;
  end;
