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

theorem Th21:
  m mod n = (n*k + m) mod n
proof
  per cases;
  suppose
A1: n > 0;
    m mod n = t implies (n*k + m) mod n = t
    proof
      assume m mod n = t;
      then consider q being Nat such that
A2:   m = n * q + t and
A3:   t < n by A1,Def2;
A0:   k is Nat & m is Nat & n is Nat & t is Nat by TARSKI:1;
      ex p be Nat st n*k + m = n*p + t & t < n
      proof
        reconsider p = k+q as Element of NAT by ORDINAL1:def 12;
        take p;
        thus thesis by A2,A3;
      end;
      hence thesis by A0,Def2;
    end;
    hence thesis;
  end;
  suppose n = 0;
    hence thesis;
  end;
end;
