reserve i, j, k, l, m, n, t for Nat;

theorem
  i-'j-'k = i-'(j+k)
proof
  per cases;
  suppose
A1: i <= j;
    hence i-'j-'k = 0-'k by Th8
      .= 0 by Th8
      .= i-'(j+k) by A1,Th8,NAT_1:12;
  end;
  suppose
A2: j <= i & i-j <= k;
    then
A3: i <= j+k by XREAL_1:20;
    i-'j = i-j by A2,XREAL_1:233;
    hence i-'j-'k = 0 by A2,Th8
      .= i-'(j+k) by A3,Th8;
  end;
  suppose
A4: j <= i & k <= i-j;
    then
A5: k+j <= i by XREAL_1:19;
    i-'j = i-j by A4,XREAL_1:233;
    hence i-'j-'k = i-j-k by A4,XREAL_1:233
      .= i-(j+k)
      .= i-'(j+k) by A5,XREAL_1:233;
  end;
end;
