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I am trying to prove that axioms A1, A2, A3, A4, M2, D hold.

so A1 :

$ a+b=b+a$ $x+x+x+x=x+x+x+x$ $4x=4x$ Am I on the right way?

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    You need to tell us what you mean by A1, etc. These are labels in your texbook, not universally accepted labels.2012-03-09

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No, you are not. In order to show that $a+b=b+a\tag{1}$ for all elements $a,b\in S$, you merely have to observe that since $x$ is the only element of $S$, the only thing that you can substitute for $a$ and $b$ in $(1)$ is $x$. When you do that, you get $x+x=x+x$; is that true? Yes, because both sides are equal to $x$ by the definition of $+$.

I’ll do one other example as an illustration. I don’t know for sure, but I suspect that your D is $a\cdot(b+c)=(a\cdot b)+(a\cdot c)\tag{2}$ for all $a,b,c\in S$. Again, the only thing in $S$ is $x$, so the only possible instance of $(2)$ to be checked is $x\cdot(x+x)=(x\cdot x)+(x\cdot x)$; is it true? On the lefthand side we have $x\cdot(x+x)=x\cdot x=x\;,$ and on the righthand side we have $(x\cdot x)+(x\cdot x)=x+x=x\;;$ these are indeed equal, so $(2)$ is true for all possible choices of $a,b$, and $c$.

One last comment: $4x$ makes no sense here. We have not defined any operation that combines natural numbers and this object $x$.

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    @David: As a notational convention, yes. But given the overall confusion, I think it important to be clear about what is really going on, which is *not* the multiplication that it might naïvely be taken as.2012-03-09
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Hint $\ $ The definitions simply transport the ring structure from the subring $\:\{0\}\subset \mathbb Z\:$ to $\{x\}$.

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Most of the required properties of a ring are immediate for the trivial ring, since any expressions on the left- or right-hand side evaluate to $x$, the unique element.

That leaves the existence of a additive identity and inverse, met by $x+x = x$ and a multiplicative identity, met by $x \cdot x = x$.