theorem Th26:
  m <= n + 1 implies A |^ (m, n + 1) = A |^ (m, n) \/ (A |^ (n + 1 ))
proof
  assume
A1: m <= n + 1;
  per cases by A1,NAT_1:8;
  suppose
A2: m <= n;
    n < n + 1 by NAT_1:13;
    then A |^ (m, n + 1) = A |^ (m, n) \/ A |^ (n + 1, n + 1) by A2,Th25;
    hence thesis by Th22;
  end;
  suppose
A3: m = n + 1;
    then
A4: m > n by NAT_1:13;
    thus A |^ (m, n + 1) = {} \/ A |^ (n + 1) by A3,Th22
      .= A |^ (m, n) \/ A |^ (n + 1) by A4,Th21;
  end;
end;
