thus Sum (digits (48,10)) = 12 by Th99; :: according to NUMBER11:def 1 :: thesis: ( 12 divides 48 & ( for m being Nat st Sum (digits (m,10)) = 12 & 12 divides m holds
48 <= m ) )
48 = 4 * 12 ;
hence 12 divides 48 by INT_1:def 3; :: thesis: for m being Nat st Sum (digits (m,10)) = 12 & 12 divides m holds
48 <= m
let m be Nat; :: thesis: ( Sum (digits (m,10)) = 12 & 12 divides m implies 48 <= m )
assume A1: ( Sum (digits (m,10)) = 12 & 12 divides m ) ; :: thesis: 48 <= m
then consider j being Nat such that
A2: m = 12 * j by NAT_D:def 3;
assume m < 48 ; :: thesis: contradiction
then 12 * j < 12 * 4 by A2;
then j < 3 + 1 by XREAL_1:64;
then j <= 3 by NAT_1:9;
then not not j = 0 & ... & not j = 3 ;

per cases then ( j = 0 or j = 1 or j = 2 or j = 3 ) ;

suppose j = 0 ; :: thesis: contradiction

then Sum (digits (m,10)) = 0 by A2, Th6;
hence contradiction by A1; :: thesis: verum

end;

suppose j = 1 ; :: thesis: contradiction

then Sum (digits (m,10)) = 3 by A2, Th93;
hence contradiction by A1; :: thesis: verum

end;

suppose j = 2 ; :: thesis: contradiction

then Sum (digits (m,10)) = 6 by A2, Th95;
hence contradiction by A1; :: thesis: verum

end;

suppose j = 3 ; :: thesis: contradiction

then Sum (digits (m,10)) = 9 by A2, Th97;
hence contradiction by A1; :: thesis: verum

end;

end;