How to negate $(a=1 \text{ and } b=n) \text{ or } (a=n \text{ and } b=1)$ to get $1<a<n \text { and } 1<b<n$?

Question

n>1 is composite if and only if it can be written as a product $n=ab$ of integers $a$ and $b$ such that $1<a<n$ and $1<b<n$. If a prime number $n$ is the product of two positive integers, $n=ab$, then either $a=1$ and $b=n$ or $a=n$ and $b=1$

I'd like to try to derive the inequality for the definition of composite number by negating the expression for the definition of a prime number:

$\begin{align} n=ab \text{ is composite } & \Rightarrow & n \text{ not prime } \\ & \Rightarrow & \neg (a=1 \wedge b=n) \vee (a=n \wedge b=1) \tag{1}\\ & \Rightarrow & (1<a \vee b<n) \wedge (a<n \vee 1<b) \\ \end{align}$

Is line 1 allowed? I have a feeling its not allowed as the negation of $p\Rightarrow q$ is not $\neg p \Rightarrow \neg q$. Even if it is not allowed, is it possible to extend the above chain of implications to arrive at $1<a<n \wedge 1<b<n$? I am stuck at the last line.

Answer 1

No, it's not allowed. And the step from line 1 to the line below it is also false. That's not what the definition states.

Also, I'm assuming that prime numbers are defined as integers $n>1$ that are not composite.

So let's break down the last sentence of the definition:

If a prime number $n$ (...)

That means: if $n>1$ can't be written as a product $n=a'b'$ of integers $a'$ and $b'$ such that $1<a'<n$ and $1<b'<n$ (...)

(...) is the product of two positive integers, $n=ab$, then either $a=1$ and $b=n$ or $a=n$ and $b=1$.

That means: (...) $n=ab$, where $a>0$ and $b>0$, then either $a=1$ and $b=n$ or $a=n$ and $b=1$.

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Proof of the last statement:

First you need to prove that $a\le n$ and $b\le n$ (which might not be trivial depending on how you defined the previous stuff, like multiplication and addition).

Then by definition,

$$\begin{align} &\; \neg \left(1<a \wedge a<n \wedge 1<b \wedge b<n\right)\\ \Rightarrow &\; 1\ge a\vee a\ge n\vee 1\ge b\vee b\ge n\\ \end{align}$$

Distributing $a>0$, $b>0$, $a\le n$ and $b\le n$ on that expression then simplifying things like $a\le 1\vee a>0$ to $a=1$, you get: $a=1\vee a=n\vee b=1\vee b=n$. This gives you four possibilities. Use $ab=n$ on each of those to show that $a=1\Rightarrow b=n$, etc., and conclude that there are only two possibilities: either $a=1$ and $b=n$ or $a=n$ and $b=1$.