reserve i, i1, i2, j, k for Nat,
  r, s for Real;

theorem Th1:
  for i, j being Nat holds j<i & i<j+j implies i mod j<>0
proof
  let i, j be Nat;
  assume that
A1: j<i and
A2: i<j+j;
A3: i-j<j+j-j by A2,XREAL_1:9;
  i-j=i-'j by A1,XREAL_1:233;
  then
A4: (i-'j) mod j=i-'j by A3,NAT_D:24;
  assume
A5: i mod j=0;
  i=i-j+j;
  then i mod j=(i-'j+j) mod j by A1,XREAL_1:233
    .=((i-'j)+ (j mod j)) mod j by NAT_D:23
    .=(i-'j+0)mod j by NAT_D:25
    .=(i-'j)mod j;
  hence contradiction by A1,A5,A4,NAT_D:36;
end;
