Let's analyze the logical proposition $b$ and its negation $\neg b$ to determine their logical values.

Proposition $b$:

$b : \forall x \in \mathbb{N}, \exists y \in \mathbb{Z} \text{ such that } x + y = 5$

This statement means that for every natural number $x$, there exists an integer $y$ such that $x + y = 5$.

Analysis of $b$:

• For any $x \in \mathbb{N}$, we need to find a $y \in \mathbb{Z}$ such that the equation $x + y = 5$ is true.
• Solving for $y$, we get $y = 5 - x$.
• Since $y = 5 - x$ will always yield an integer (because $\mathbb{Z}$ includes all integers, both positive and negative), such a $y$ will always exist for any $x \in \mathbb{N}$.

Therefore, the proposition $b$ is true ($b = T$).

Negation $\neg b$:

The negation of $b$ is:

$\neg b : \exists x \in \mathbb{N} \text{ such that } \forall y \in \mathbb{Z}, x + y \neq 5$

This means that there exists at least one natural number $x$ such that for all integers $y$, the equation $x + y = 5$ does not hold.

However, since for every $x \in \mathbb{N}$, there is a $y \in \mathbb{Z}$ such that $x + y = 5$, the negation $\neg b$ is false ($\neg b = F$).

Conclusion:

• The logical value of $b$ is true ($b = T$).
• The negation $\neg b$ is false ($\neg b = F$).

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the proposition $b$ change if we replaced $\mathbb{Z}$ with $\mathbb{N}$ for $y$?
2. What is the impact of changing the domain of $x$ from $\mathbb{N}$ to $\mathbb{Z}$ in the proposition $b$?
3. How would you negate a proposition where both $x$ and $y$ are quantified by $\forall$ instead of $\forall$ and $\exists$?
4. What does the negation of a proposition involving universal quantifiers tell us about the original statement?
5. Can you formulate a similar proposition where $\neg b$ would be true?

Tip: When working with logical quantifiers, always consider the domains of the variables carefully, as they affect the truth values and the negation of the proposition.