reserve a, b, k, n, m for Nat,
  i for Integer,
  r for Real,
  p for Rational,
  c for Complex,
  x for object,
  f for Function;

theorem
  divSeq(0,n) = NAT --> 0
proof
  set fd = divSeq(0,n);
A1: for x being object st x in dom fd holds fd.x = g.x
  proof
    defpred P[Nat] means fd.$1 = 0;
    let x be object;
    assume x in dom fd;
    then reconsider x as Element of NAT;
A2: for x being Nat holds P[x] implies P[x+1]
    proof
      let x be Nat;
      assume
A3:   P[x];
      per cases;
      suppose
A4:     x = 0;
        fd.1 = n div (0 mod n) by Def2
          .= n div 0;
        hence thesis by A4;
      end;
      suppose
        0 <> x;
        hence thesis by A3,Th17,NAT_1:11;
      end;
    end;
    fd.0 = 0 div n by Def2
      .= 0;
    then
A5: P[0];
    for n being Nat holds P[n] from NAT_1:sch 2(A5,A2);
    then fd.x = 0;
    hence thesis by FUNCOP_1:7;
  end;
  dom fd = NAT by FUNCT_2:def 1;
  hence thesis by A1;
end;
