시험시간에 이 문제에서 ZF공리는 사용해도 된다고 하셨는데요,
ZF공리를 사용하면 (i)과 (iv)만 가지고 (ii)와 (iii)을 증명할 수 있습니다.

(ii) If x∈a and x=a, by axiom of equality, we have x∈x.
     By axiom of pair, {x,x}={x} is a set.
    Then, by axiom of foundation, there is an element of {x}, b such that b∩{x}=0. Clearly b=x.
     However, the intersection x∩{x} contains x, which is a contradiction.

(iii) By (ii), clearly x is not y and y is not z.
      Suppose that x=z. Then, y∈z and z=x. By axiom of equality, y∈x. Note that x∈y.
      By axiom of pair, S={x,y} is a set. By axiom of foundation, x∩S=0 or y∩S=0.
      If x∩S=0, y∈x and y∈S. If y∩S=0, x∈y and x∈S, and both are contradictions.
      Therefore, x is not equal to z.
      
      Now consider T={x,z}, which is a set by axiom of pair.
      By (iv), there exists t∈T such that
      <t∈x or t=x> and <t∈z or t=z>. Clearly, t=x or t=z.
      If t=z, we have z∈x, since z is not equal to x.
      By axiom of pair, {x,y} is a set and {z,z}={z} is a set.
      By axiom of pair, {{x,y},{z}} is a set.
      By axiom of union, V=U{{x,y},{z}}={x,y,z} is a set.
      We have x∈y∈z∈x. By axiom of foundation, x∩V=0 or y∩V=0 or z∩V=0, but all the three yields a
      contradiction.
      Thus, t=x. Since x is not equal to z, x∈z.

그렇다면 (ii)와 (iii)을 정의에서 제외시키는 것이 좀 더 좋은 정의가 아닐까요?