
theorem Th3:
  for i, j being Nat st i > i -' 1 + j holds j = 0
proof
  let i, j be Nat;
  assume i > i-'1+j;
  then i-j+j > i-'1+j;
  then A2: i-j > i-'1 by XREAL_1:6;
  i-j >= 1
  proof
    assume A4: i-j < 1;
    i-j+j > 0+j by A2, XREAL_1:6;
    then i-j is Nat by NAT_1:21;
    hence thesis by A2, A4, NAT_1:14;
  end;
  then i-j = i-'j by NAT_D:39;
  hence j = 0 by A2, NAT_1:14, NAT_D:41;
end;
