# Proof in FOL with no CQ rule

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

• What "CQ" means? – Mauro ALLEGRANZA Mar 1 at 7:30
• @MauroALLEGRANZA Change of Quantifier. The rules of Quantifier Duality. $$\lnot\exists x~\varphi~\iff~ \forall x~\lnot\varphi\\\lnot\forall x~\varphi~\iff~ \exists x~\lnot\varphi$$ – Graham Kemp Mar 2 at 0:04

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

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

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...