thus Sum (digits (195,10)) = 15 by Th126; :: according to NUMBER11:def 1 :: thesis: ( 15 divides 195 & ( for m being Nat st Sum (digits (m,10)) = 15 & 15 divides m holds
195 <= m ) )

195 = 13 * 15 ;
hence 15 divides 195 by INT_1:def 3; :: thesis: for m being Nat st Sum (digits (m,10)) = 15 & 15 divides m holds
195 <= m

let m be Nat; :: thesis: ( Sum (digits (m,10)) = 15 & 15 divides m implies 195 <= m )
assume A1: ( Sum (digits (m,10)) = 15 & 15 divides m ) ; :: thesis: 195 <= m
then consider j being Nat such that
A2: m = 15 * j by NAT_D:def 3;
assume m < 195 ; :: thesis: contradiction
then 15 * j < 15 * 13 by A2;
then j < 12 + 1 by XREAL_1:64;
then j <= 12 by NAT_1:9;
then not not j = 0 & ... & not j = 12 ;
per cases then ( j = 0 or j = 1 or j = 2 or j = 3 or j = 4 or j = 5 or j = 6 or j = 7 or j = 8 or j = 9 or j = 10 or j = 11 or j = 12 ) ;
suppose j = 0 ; :: thesis: contradiction
end;
suppose j = 1 ; :: thesis: contradiction
end;
suppose j = 2 ; :: thesis: contradiction
end;
suppose j = 3 ; :: thesis: contradiction
end;
suppose j = 4 ; :: thesis: contradiction
end;
suppose j = 5 ; :: thesis: contradiction
end;
suppose j = 6 ; :: thesis: contradiction
end;
suppose j = 7 ; :: thesis: contradiction
end;
suppose j = 8 ; :: thesis: contradiction
end;
suppose j = 9 ; :: thesis: contradiction
end;
suppose j = 10 ; :: thesis: contradiction
end;
suppose j = 11 ; :: thesis: contradiction
end;
suppose j = 12 ; :: thesis: contradiction
end;
end;