
theorem Th1:
  for n,k being Nat st n|^|^k = 0 holds n = 0
  proof
    let n,k be Nat;
    assume A1: n|^|^k = 0;
    then k <> 0 by ORDINAL5:13;
    then consider j being Nat such that
    A2: k=j+1 by NAT_1:6;
    n|^|^(j+1) = n|^|^(Segm (j+1))
    .= n|^|^(succ Segm j) by NAT_1:38
    .= exp(n,n|^|^j) by ORDINAL5:14
    .= n|^(n|^|^j);
    hence n=0 by A1,A2;
  end;
