theorem Th2:
  3 <= i implies i mod (i-'1) = 1
proof
  assume
A1: 3 <= i;
  then
A2: i-'1 = i-1 by XREAL_1:233,XXREAL_0:2;
  3-'1 = 2+1-'1 .= 2 by NAT_D:34;
  then 2 <= i-'1 by A1,NAT_D:42;
  then 1 < i-1 by A2,XXREAL_0:2;
  then 1+i < (i-1)+i by XREAL_1:8;
  then 1+i-1 < (i-1)+i-1 by XREAL_1:14;
  then
A3: i < (i-1) + (i-1);
  i-'1 <= i by NAT_D:35;
  hence i mod (i-'1) = i - (i-1) by A2,A3,JORDAN4:2
    .= 1;
end;
