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!!
On
HINT
It's always good to think about this informally, and then try to formalize that. Thinking about it informally will clarify your all-important proof strategy.
OK, so you want to prove:
$$\forall x\forall y(S(x, y) \Rightarrow \neg S(y, x))$$
This means that you consider and $x$ and $y$, assume that $S(x,y)$, and then try to show $\neg S(y,x)$.
And for the latter, since it is a negation, probably a proof by contradiction will be good, i.e. assume that you do have $S(y,x)$, and show that that leads to a contradiction.
Hint: Assume $S(x,y)$ and $S(y,x)$. Then $S(x,x)$ by the first premise, contradicting the second premise.