Proof in FOL with no CQ rule

Question

I had the first completed but when doing its converse, since there is no CQ rule, I do not know where to proceed. Please help

Answer 1

1. $\lnot \exists x (P(x) \land Q(x))$ --- premise

2. $P(a)$ --- assumed [a]

3. $Q(a)$ --- assumed [b]

4. $\exists x (P(x) \land Q(x))$

5. $\bot$

6. $\lnot Q(a)$

7. $P(a) \to \lnot Q(a)$

8. $\forall x (P(x) \to \lnot Q(x))$

Answer 2

As usual the key is to work backwards.

To introduce the universal, assume an arbitrary variable, $a$, then derive $P(a)\to\lnot Q(a)$.

To introduce that conditional, assume $P(a)$ then derive $\lnot Q(a)$.

To introduce that negation, assume $Q(a)$, then derive a contradiction.

To derive that contradiction under the assumptions of $\lnot\exists x~(P(x)\land Q(x))$, $P(a)$, and $Q(a)$, well...