Meaning of $∃y∃z : \text{times}(y,z) = x \land (\neg (y=1) \land \neg (z=1))$

Question

I have the following question:

$P(x) \iff ∃y∃z : \text{times}(y,z) = x \land (\neg (y=1) \land \neg (z=1))$

$\text{times}$ is the multiplication function
The world is a world of natural numbers

Question: Whats the meaning of predicate $P$ above?

Answer: $x$ is a composite number (not prime and not equal to $1$)

I'm new to this, and I tried really hard to understand the question, however I can't get it. Whats the steps/way to solve a question like this?

Thanks

Answer 1

You have to "unwind" the formula :

$P(x) \leftrightarrow ∃y∃z[x=\times(y,z)∧(¬(y=1)∧¬(z=1))]$.

The "universe" is the set $\mathbb N$ of natural numbers; thus $P(n)$ holds exactly when the RHS is satisfied by $n$.

This amount to : $n=\times(y,z)∧(¬(y=1)∧¬(z=1))$ for some $y$ and some $z$ in the universe, i.e. :

for some $j,k \in \mathbb N : \, (n=j \times k)∧(j \ne 1)∧(k \ne 1)$.

Now it is quite easy to "see" that $P(n)$ holds exactly when $n$ is a composite number, i.e. it is not a prime, because it is expressible as the product of two numbers both different from $1$.

Answer 2

We know that $x$ is a composite number if it is not a prime and not equal to $1$. Your predicate $P(x)$ reads as follows:

There exist $y$ and $z$ such that $y \cdot z = x$ with $y \not = 1$ and $z \not = 1$.

We want to show that this holds if and only if $x$ is a composite number. Therefore, assume $x = 1$, then since $y,z$ are natural numbers, we have $y=z=1$ but this contradicts $y,z \not = 1$. Now assume $x$ is a prime, then we know that one of $y,z$ equals $x$ and the other equals $1$, because a prime has no other divisors than one and itself. So again, we have the same contradiction. This shows that only composite numbers $x$ can work for $P(x)$.

We still have to show that $P(x)$ holds for all composite numbers, but this is easy since we know that for every composite number $x$ exist $a,b \not = 1$ with $a \cdot b = x$.