reserve k for Nat;
reserve p for Prime;

theorem Ttool127a:
  p < 127 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 or p = 113
  proof
    assume p < 127;
    then 1+1 < p+1 & p < 126+1 by XREAL_1:6,INT_2:def 4;
    then per cases by NAT_1:13;
    suppose 2 <= p < 113;
      hence thesis by Ttool113a;
    end;
    suppose 113 <= p & p <= 120;
      then 113 <= p <= 113+1 or 114 <= p <= 114+1 or 115 <= p <= 115+1 or 
        116 <= p <= 116+1 or 117 <= p <= 117+1 or 118 <= p <= 118+1 or 
        119 <= p <= 119+1;
      then p = 113 by XPRIMES0:114,115,116,117,118,119,120,NAT_1:9;
      hence thesis;
    end;
    suppose 120 <= p <= 120+1 or 121 <= p <= 121+1 or 122 <= p <= 122+1 or
      123 <= p <= 123+1 or 124 <= p <= 124+1 or 125 <= p <= 125+1;
      then contradiction by XPRIMES0:120,121,122,123,124,125,126,NAT_1:9;
      hence thesis;
    end;
  end;
