
theorem Th21:
  for L being Field, i,j being Integer, x being Element of L st x
  <> 0.L holds pow(x,i) * pow(x,j) = pow(x,i+j)
proof
  let L be Field;
  let i,j be Integer;
  let x be Element of L;
  defpred P[Integer] means for i being Integer holds pow(x, i+$1) = pow(x, i)
  * pow(x, $1);
  assume
A1: x <> 0.L;
A2: for j being Integer holds P[j] implies P[j - 1] & P[j + 1]
  proof
    let j be Integer;
    assume
A3: for i being Integer holds pow(x, i+j) = pow(x, i) * pow(x, j);
    thus for i being Integer holds pow(x, i + (j - 1)) = pow(x, i) * pow(x, j
    - 1)
    proof
      let i be Integer;
      thus pow(x, i + (j - 1)) = pow(x, (i - 1) + j)
        .= pow(x, i - 1) * pow(x, j) by A3
        .= (pow(x, i) * pow(x, -1)) * pow(x, j) by A1,Th20
        .= pow(x, i) * (pow(x, -1) * pow(x, j)) by GROUP_1:def 3
        .= pow(x, i) * pow(x, j + (-1)) by A3
        .= pow(x, i) * pow(x, j - 1);
    end;
    let i be Integer;
    thus pow(x, i + (j + 1)) = pow(x, (i + 1) + j)
      .= pow(x, i + 1) * pow(x, j) by A3
      .= (pow(x, i) * pow(x, 1)) * pow(x, j) by A1,Th19
      .= pow(x, i) * (pow(x, 1) * pow(x, j)) by GROUP_1:def 3
      .= pow(x, i) * pow(x, j + 1) by A3;
  end;
A4: P[0]
  proof
    let i be Integer;
    thus pow(x, i+0) = pow(x, i) * 1.L
      .= pow(x, i) * pow(x, 0) by Th13;
  end;
  for j being Integer holds P[j] from INT_1:sch 4(A4,A2);
  hence thesis;
end;
