reserve k for Nat;
reserve p for Prime;

theorem Ttool113a:
  p < 113 implies
  p = 2 or p = 3 or p = 5 or p = 7 or p = 11 or p = 13 or p = 17 or 
  p = 19 or p = 23 or p = 29 or p = 31 or p = 37 or p = 41 or p = 43 or 
  p = 47 or p = 53 or p = 59 or p = 61 or p = 67 or p = 71 or p = 73 or 
  p = 79 or p = 83 or p = 89 or p = 97 or p = 101 or p = 103 or p = 107 or 
  p = 109
  proof
    assume p < 113;
    then 1+1 < p+1 & p < 112+1 by XREAL_1:6,INT_2:def 4;
    then per cases by NAT_1:13;
    suppose 2 <= p < 109;
      hence thesis by Ttool109a;
    end;
    suppose 109 <= p <= 109+1 or 110 <= p <= 110+1 or 111 <= p <= 111+1;
      then p = 109 by XPRIMES0:110,111,112,NAT_1:9;
      hence thesis;
    end;
  end;
