let n be Nat; :: thesis: ( not 5 divides n iff ex k being Nat st
( n = (5 * k) + 1 or n = (5 * k) + 2 or n = (5 * k) + 3 or n = (5 * k) + 4 ) )
consider K being Nat such that
A1: ( n = 5 * K or n = (5 * K) + 1 or n = (5 * K) + 2 or n = (5 * K) + 3 or n = (5 * K) + 4 ) by Th25;
thus ( not 5 divides n implies ex k being Nat st
( n = (5 * k) + 1 or n = (5 * k) + 2 or n = (5 * k) + 3 or n = (5 * k) + 4 ) ) by A1; :: thesis: ( ex k being Nat st
( n = (5 * k) + 1 or n = (5 * k) + 2 or n = (5 * k) + 3 or n = (5 * k) + 4 ) implies not 5 divides n )
given k being Nat such that A2: ( n = (5 * k) + 1 or n = (5 * k) + 2 or n = (5 * k) + 3 or n = (5 * k) + 4 ) ; :: thesis: not 5 divides n
given t being Nat such that A3: n = 5 * t ; :: according to NAT_D:def 3 :: thesis: contradiction

per cases ( n = (5 * k) + 1 or n = (5 * k) + 2 or n = (5 * k) + 3 or n = (5 * k) + 4 ) by A2;

suppose n = (5 * k) + 1 ; :: thesis: contradiction

then 1 / 5 = t - k by A3;
hence contradiction by NAT_D:33; :: thesis: verum

end;

suppose n = (5 * k) + 2 ; :: thesis: contradiction

then 2 / 5 = t - k by A3;
hence contradiction by NAT_D:33; :: thesis: verum

end;

suppose n = (5 * k) + 3 ; :: thesis: contradiction

then 3 / 5 = t - k by A3;
hence contradiction by NAT_D:33; :: thesis: verum

end;

suppose n = (5 * k) + 4 ; :: thesis: contradiction

then 4 / 5 = t - k by A3;
hence contradiction by NAT_D:33; :: thesis: verum

end;

end;