let n be Nat; :: thesis: ( not 8 divides n iff ex k being Nat st
( n = (8 * k) + 1 or n = (8 * k) + 2 or n = (8 * k) + 3 or n = (8 * k) + 4 or n = (8 * k) + 5 or n = (8 * k) + 6 or n = (8 * k) + 7 ) )

consider K being Nat such that
A1: ( n = 8 * K or n = (8 * K) + 1 or n = (8 * K) + 2 or n = (8 * K) + 3 or n = (8 * K) + 4 or n = (8 * K) + 5 or n = (8 * K) + 6 or n = (8 * K) + 7 ) by Th28;
thus ( not 8 divides n implies ex k being Nat st
( n = (8 * k) + 1 or n = (8 * k) + 2 or n = (8 * k) + 3 or n = (8 * k) + 4 or n = (8 * k) + 5 or n = (8 * k) + 6 or n = (8 * k) + 7 ) ) by A1; :: thesis: ( ex k being Nat st
( n = (8 * k) + 1 or n = (8 * k) + 2 or n = (8 * k) + 3 or n = (8 * k) + 4 or n = (8 * k) + 5 or n = (8 * k) + 6 or n = (8 * k) + 7 ) implies not 8 divides n )

given k being Nat such that A2: ( n = (8 * k) + 1 or n = (8 * k) + 2 or n = (8 * k) + 3 or n = (8 * k) + 4 or n = (8 * k) + 5 or n = (8 * k) + 6 or n = (8 * k) + 7 ) ; :: thesis: not 8 divides n
given t being Nat such that A3: n = 8 * t ; :: according to NAT_D:def 3 :: thesis: contradiction
per cases ( n = (8 * k) + 1 or n = (8 * k) + 2 or n = (8 * k) + 3 or n = (8 * k) + 4 or n = (8 * k) + 5 or n = (8 * k) + 6 or n = (8 * k) + 7 ) by A2;
suppose n = (8 * k) + 1 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 2 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 3 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 4 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 5 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 6 ; :: thesis: contradiction
end;
suppose n = (8 * k) + 7 ; :: thesis: contradiction
end;
end;