How to prove this sequent by natural deduction? How do I prove
$$\forall x\forall y\forall z(S(x, y)\land S(y, z) \Rightarrow S(x, z)), \forall x\neg S(x, x) \vdash \forall x\forall y(S(x, y) \Rightarrow \neg S(y, x)).$$
by natural deduction?
1 $\quad \forall x\forall y\forall z(S(x, y)\land S(y, z) \Rightarrow S(x, z))\quad \text{premise}$ 
2 $\quad\forall x\neg S(x, x) \quad\text{premise}$
I don't know what's the next step, replace $x$ by some term?
I got it!!


Hint: Assume $S(x,y)$ and $S(y,x)$. Then $S(x,x)$ by the first premise, contradicting the second premise.
