这是一份oxford牛津大学作业代写的成功案例
Consider the proposition
$$
A=(\neg P \supset Q) \supset(\neg R \supset S) .
$$
First, we eliminate $\supset$ using the fact that $(\neg B \vee C)$ is equivalent to $(B \supset C)$. We get
$$
(\neg(\neg \neg P \vee Q)) \vee(\neg \neg R \vee S) .
$$
Then, we put this proposition in NNF. We obtain
$$
(\neg P \wedge \neg Q) \vee(R \vee S)
$$
Using distributivity we obtain
$$
(\neg P \vee R \vee S) \wedge(\neg Q \vee R \vee S)
$$
Oxford COURSE NOTES ：
Show that the infinite sequent $\Gamma \rightarrow \Delta$ where
$$
\Gamma=
$$
and
$$
\left.\Delta=<\left(P_{1} \supset Q\right)\right\rangle $$ is falsifiable. (ii) Prove that for every $i>0$, the sequent $\Gamma \rightarrow \Delta^{\prime}$, where $\Gamma$ is as above and $\Delta^{\prime}=<\left(P_{0} \supset P_{i}\right)>$ is provable.