theorem Th5:
  n >= 1 implies (0.K) |^ n = 0.K
  proof
    set a1 = 0.K;
    assume A2: n >= 1;
    n - 1 in NAT by A2,INT_1:5;
    then consider n1 be Nat such that
A3: n1 = n - 1;
    a1 |^n = a1 |^(n1+1) by A3
    .= a1|^n1 * a1 by EC_PF_1:24;
    hence a1 |^n = 0.K;
  end;
