reserve a,b,p,k,l,m,n,s,h,i,j,t,i1,i2 for natural Number;

theorem Th27:
  i < j implies i div j = 0
proof
  assume
A1: i<j;
  per cases by Def1;
  suppose ex j1 being Nat st i=j*(i div j)+j1 & j1<j;
    then consider j1 being Nat such that
A2: i=j*(i div j)+j1 and j1<j;
    assume i div j<>0;
    then consider k be Nat such that
A3: i div j = k+1 by NAT_1:6;
    i = j + (j*k+j1) by A2,A3;
    hence contradiction by A1,NAT_1:11;
  end;
  suppose i div j=0 & j=0;
    hence thesis;
  end;
end;
