Let's analyze the statement and see if it's true. The problem can be broken down as follows:

Given:

• $U = \mathbb{N}$ (the set of natural numbers).
• $P(x, y, z)$ is the predicate $xy - z \leq x^2 - 5$.

We want to verify if the implication is true:

$\forall x \, \exists y \, \forall z \, P(x, y, z) \rightarrow \forall z \, P(5, 4, z)$

Step 1: Analyze $\forall x \, \exists y \, \forall z \, P(x, y, z)$

This means that for every natural number $x$, there exists a natural number $y$ such that for all natural numbers $z$, the predicate $xy - z \leq x^2 - 5$ holds true.

Let's express this for arbitrary $x$:

$xy - z \leq x^2 - 5 \quad \text{for all } z$

This inequality can be rearranged as:

$xy \leq x^2 - 5 + z$

Step 2: Analyze $\forall z \, P(5, 4, z)$

Substituting $x = 5$ and $y = 4$ into the predicate, we get:

$5 \cdot 4 - z \leq 5^2 - 5$

Simplifying, this becomes:

$20 - z \leq 25 - 5$

$20 - z \leq 20$

This simplifies further to:

$-z \leq 0 \quad \text{or} \quad z \geq 0$

Since $z$ is a natural number ($z \in \mathbb{N}$), this inequality is always true.

Step 3: Determine if the implication holds

Now, let's review the original implication:

$\forall x \, \exists y \, \forall z \, P(x, y, z) \rightarrow \forall z \, P(5, 4, z)$

From Step 2, we established that $\forall z \, P(5, 4, z)$ holds true. Since the right side of the implication is always true, the implication as a whole is also true, regardless of the truth value of the left side.

Conclusion

The statement:

$\forall x \, \exists y \, \forall z \, P(x, y, z) \rightarrow \forall z \, P(5, 4, z)$

is true.

Would you like more details on any part of the explanation?

Here are some related questions you might consider:

1. What if $U = \mathbb{Z}$ instead of $U = \mathbb{N}$?
2. How would the analysis change if $P(x, y, z)$ was a different predicate?
3. Can we find explicit values of $y$ that satisfy $\forall x \, \exists y \, \forall z \, P(x, y, z)$?
4. How does the truth value change if $x = 0$ is included in $U$?
5. What would happen if $\exists y$ and $\forall z$ were swapped in the original quantifier expression?

Tip: When dealing with logical implications, if the consequent (right side of the implication) is always true, the entire implication will also always be true, regardless of the antecedent (left side).