let i be non zero Integer; :: thesis: sin is (2 * PI) * i -periodic
i in INT by INT_1:def 2;
then consider k being Element of NAT such that
a1: ( i = k or i = - k ) by INT_1:def 1;

per cases ( i = k or i = - k ) by a1;

suppose i = k ; :: thesis: sin is (2 * PI) * i -periodic

then ex m being Nat st i = m + 1 by NAT_1:6;
hence sin is (2 * PI) * i -periodic by Lmsin; :: thesis: verum

end;

suppose a2: i = - k ; :: thesis: sin is (2 * PI) * i -periodic

then consider m being Nat such that
a3: - i = m + 1 by NAT_1:6;
sin is (2 * PI) * (m + 1) -periodic by Lmsin;
then sin is - ((2 * PI) * (m + 1)) -periodic by Th14;
hence sin is (2 * PI) * i -periodic by a2, a3; :: thesis: verum

end;

end;