reserve k for Nat;
reserve p for Prime;

theorem Ttool149a:
  p < 149 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 or p = 127 or p = 131 or p = 137 or p = 139
  proof
    assume p < 149;
    then 1+1 < p+1 & p < 148+1 by XREAL_1:6,INT_2:def 4;
    then per cases by NAT_1:13;
    suppose 2 <= p < 139;
      hence thesis by Ttool139a;
    end;
    suppose 139 <= p <= 139+1 or 140 <= p <= 140+1 or 141 <= p <= 141+1 or 
      142 <= p <= 142+1 or 143 <= p <= 143+1 or 144 <= p <= 144+1 or 
      145 <= p <= 145+1 or 146 <= p <= 146+1 or 147 <= p <= 147+1;
      then p = 139 by XPRIMES0:140,141,142,143,144,145,146,147,148,NAT_1:9;
      hence thesis;
    end;
  end;
