A property of semiring Definition 1.
Let $(R, +, \cdot)$ be a semiring equipped with a relation $\leq$ such that the relation is defined as: for all $x, y \in R$, $x\leq y$ if and only if $x+a=y$ (or $a+x=y$) for some $a\in R.$
Proposition. Let $R$ be a multiplicatively idempotent and additively cancellative semiring with additive identity. Then $x\leq 2x\leq 3x\leq 4x\leq 5x\leq 6x\leq 7x\leq 8x... ~\text{for all} ~ x\in R$.
Proof.
Considering $R$ to be an orderable semiring under the partial order relation $\leq$ (Definition 1) over $R$ and $x\leq y$ for all $x, y\in R$, we can write $a+x=y$ for some $a\in R$ or $x\leq a+x$. Without loss of generality, we can put $a=x$, which implies that $x\leq x+x$ or $x\leq 2x$. But $R$ is additively cancellative, hence we can add $x$ both sides of the inequality $x\leq 2x$ to obtain $2x\leq 3x$. Similarly we can keep adding $x$ to each subsequent inequality to obtain $nx\leq (n+1)x$ in general for all $n\geq 1$. Thus we have $x\leq 2x\leq 3x\leq 4x\leq 5x\leq 6x\leq 7x\leq 8x... ~\text{for all} ~ x\in R$
My problem in the proof is stated below:

Without loss of generality, we can put $a=x$ . Am I doing right by putting $x=a$?.

Any alternative proofs are welcome.
 A: wlog
The phrase "without loss of generality" isn't used the way you are applying it.
It is used to mean "while it might look like this is narrower than necessary, it in fact is sufficient to prove this." As Wikipedia puts it: "The term is used to indicate the assumption that follows is chosen arbitrarily, narrowing the premise to a particular case, but does not affect the validity of the proof in general."
The simplest example I can think of is this, you might be proving something about rational numbers, and you want $\frac{a}{b}\in \mathbb Q$. You can say "without loss of generality, $\frac{a}{b}$ is in lowest terms."
Conducting the proof
I would rather phrase what you have done this way:

In particular, when $a=x$ it follows that $x\leq 2x$.

Then to prove the chain of inequalities, it would be natural to couch it in terms so that the principle of induction can take hold. For example, say $P(n)$ means $x\leq 2x\leq 3x\leq ...\leq nx$ whenever $n$ is a positive integer. By establishing $x\leq 2x$ you've got a base case that holds. You may, if you wish, back up a step to note that $0+x=x$ implies $0x\leq x$ to get a base case at $1$ as well.
Now you suppose $P(n)$ is true and prove $P(n+1)$ must follow.

Suppose that $P(n)$ holds. Then since $nx+x=(n+1)x$ implies $n\leq (n+1)x$, we may append $(n+1)x$ to the end of the chain $P(n)$ provided to conclude that $P(n+1)$ holds.  By the principle of induction we may conclude that $P(n)$ holds for all positive integers $n$.  Therefore the entire chain of inequalities holds.

What about the other hypotheses?
I can't see that they're terribly useful. It seems to me you can prove the chain of inequalities using just the definition of $\leq$.
Additive cancellativity would allow you to deduce $(n-1)x\leq nx$ from $nx\leq (n+1)x$ by cancelling $x$ from $(n-1)x+x+x=nx+x$, but that's the wrong direction, right? The multiplicative idempotence also seems immaterial. Perhaps they are just included to make the preorder $\leq$ nicer.
