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

theorem
  k divides n & 1 <= n & 1 <= i & i <= k implies (n -' i) div k = (n div k) - 1
proof
  assume that
A1: k divides n and
A2: 1 <= n and
A3: 1 <= i and
A4: i <= k;
A5: k-'i < k by A3,A4,Th9;
A6: k <= n by A1,A2,NAT_D:7;
  then i + k <= k + n by A4,XREAL_1:7;
  then
A7: i <= n by XREAL_1:6;
  n div k > 0
  proof
    assume not n div k > 0;
    then n div k = 0;
    hence contradiction by A3,A4,A6,Th12;
  end;
  then n div k >= 0 + 1 by NAT_1:13;
  then
A8: (n div k)-'1 = (n div k) - 1 by XREAL_1:233;
  n = k * (n div k) by A1,NAT_D:3;
  then n -' i = k * (n div k) - k * 1 + k - i by A7,XREAL_1:233
    .= k * ((n div k)-'1) + (k - i) by A8
    .= k * ((n div k)-'1) + (k-'i) by A4,XREAL_1:233;
  hence thesis by A8,A5,NAT_D:def 1;
end;
