theorem
 l <> 0 implies 0 #Z l = 0 proof assume
A1: l <> 0;
  per cases by A1;
  suppose
A2:  l > 0;
    then reconsider l as Element of NAT by INT_1:3;
    l >= 0+1 by A2,NAT_1:13;
    then
A3:   0 |^ l = 0 by NEWTON:11;
    |.l.| = l by ABSVALUE:def 1;
   hence thesis by A3,Def3;
  end;
  suppose
A4:  l < 0;
    then reconsider k = -l as Element of NAT by INT_1:3;
    k > -0 by A4;
    then k >= 0+1 by NAT_1:13;
    then
A5:   0 |^ k = 0 by NEWTON:11;
    |.l.| = k by A4,ABSVALUE:def 1;
    then (0 |^ |.l.|)" = 0 by A5;
   hence thesis by A4,Def3;
  end;
 end;
