Proof by contradiction structure for statement of the form $P \implies \exists x \in X: Q(x) \wedge R(x)$

Question

For a proof by contradiction, we suppose $P \wedge \forall x \in X: \neg Q(x) \lor \neg R(x)$.

Is the following proof outline sound?

We show that for a particular $a \in X$, chosen and specified,

1. Assume $P \wedge \neg Qa$ and reach (if any) a contra $A \wedge \neg A$,

2. Assume $P \wedge \neg Ra$ and reach (if any) a contra $A \wedge \neg A$

while noting that the contra in 1. and 2. have been achieved with the chosen $a \in X$.

The issue that I have is that I’m specifying a particular $a \in X$ a-priori, rather then for any $x \in X$. Is this enough since we have universal quantification?

How I’m understanding it, you’re assuming that some $A$ holds when the negation is true for all $x \in X$, however you have constructed one $a \in X$ for which we can conclude $\neg A$.

Many thanks in advance.

Answer 1

1. Assume $P$

2. Assume $\forall x\in X: [\neg Q(x) \lor \neg R(x)]$

3. Obtain contradiction $A \land \neg A$

4. Conclude $\neg\forall x \in X: [\neg Q(x) \lor \neg R(x)]$ (discharging 2)

5. $\exists x\in X: [ Q(x) \land R(x)]$ (from 4)

6. Conclude $P \implies \exists x\in X: [ Q(x) \land R(x)]$ (discharging 1)