Clarifying then negating a sentence

Question

There is a prime number $p$ such that $P(p, q)$ is true for all prime numbers $q = p$.

Notice that the given statement goes "there is a prime number $p$" followed by "$P(p, q)$ is true" and then followed by "for all prime numbers $q = p$". However, when writing it in math symbols, I put "$\forall q$" before "$P(p, q)$ is true", correct?

Given the information above, is this negation correct?

• Let $\mathbb{P}$ be the set of prime numbers.

  $$\neg(\exists p {\in}\mathbb{P} \; \space \forall q {\in} \mathbb{P}\; \space q = p \space \text{ such that } P(p, q))$$

  $$\forall p {\in} \mathbb{P}\; \space \exists q {\in} \mathbb{P}\; \space q = p \space \text{ such that } \neg P(p, q)$$

  For all prime numbers $p$, there is a prime number $q$ equal to p, for which $P(p, q)$ is false.

Answer 1

Your rectification of the hanging quantifier and final answer are both good, however your translations and intermediate step are incorrect.

There is a prime number $p$ such that $P(p, q)$ is true for all prime numbers $q = p$.

$$\neg(\exists p {\in}\mathbb{P} \; \space \forall q {\in} \mathbb{P}\; \space q = p \space \textbf{ such that } P(p, q))$$

In this intermediate step, what you've written means $$\neg\Big(\exists p {\in}\mathbb{P} \; \forall q {\in} \mathbb{P}\; \big(q = p \; \land \; P(p, q) \big)\Big),$$ whereas the correct negation is $$\neg\Big(\exists p {\in}\mathbb{P} \; \forall q {\in} \mathbb{P}\; \big(q = p \,\to\, P(p, q) \big)\Big);$$ translating this clunkily to make clear how your placement of "such that" is wrong:

• "it is not that there exists a prime $p$ such that for each prime $q$ such that $q=p,$ $P(p,q)$."

$$\forall p {\in} \mathbb{P}\; \space \exists q {\in} \mathbb{P}\; \space q = p \space \textbf{ such that } \neg P(p, q)$$

This final step, though, is technically correct; however, this if you want to spell out the "such that"s, then it is more consistent to write $$\forall p {\in} \mathbb{P}\; \space \exists q {\in} \mathbb{P}\; \textbf{ such that } \space q = p \space \textbf{ such that } \space \neg P(p, q);$$ however, $$\forall p {\in} \mathbb{P}\; \space \exists q {\in} \mathbb{P}\; \textbf{ such that } \space q = p \space \textbf{ and } \space \neg P(p, q),$$ is more natural, and $$\forall p {\in} \mathbb{P}\; \space \exists q {\in} \mathbb{P}\; \big(q = p \;\land\; \neg P(p, q)\big)$$ is best.