for n being Element of NAT st 1 < n & n * n <= 113 & n is prime holds
not n divides 113

proof

let n be Element of NAT ; :: thesis: ( 1 < n & n * n <= 113 & n is prime implies not n divides 113 )
113 = (2 * 56) + 1 ;
then A1: not 2 divides 113 ;
113 = (3 * 37) + 2 ;
then A2: not 3 divides 113 by NAT_4:9;
113 = (5 * 22) + 3 ;
then A3: not 5 divides 113 by NAT_4:9;
113 = (7 * 16) + 1 ;
then A4: not 7 divides 113 by NAT_4:9;
113 = (11 * 10) + 3 ;
then A5: not 11 divides 113 by NAT_4:9;
113 = (13 * 8) + 9 ;
then A6: not 13 divides 113 by NAT_4:9;
113 = (17 * 6) + 11 ;
then A7: not 17 divides 113 by NAT_4:9;
113 = (19 * 5) + 18 ;
then A8: not 19 divides 113 by NAT_4:9;
113 = (23 * 4) + 21 ;
then not 23 divides 113 by NAT_4:9;
hence ( 1 < n & n * n <= 113 & n is prime implies not n divides 113 ) by A1, A2, A3, A4, A5, A6, A7, A8, NAT_4:62; :: thesis: verum

end;

hence 113 is prime by NAT_4:14; :: thesis: verum