Are addition and multiplication on naturals algebraically distinguishable?

Question

Suppose (N, +) and (N, *) are the structures of addition and multiplication on N, the natural numbers with 0. Let S be the set of equational identities that hold in (N, +), and let T be the set of equational identities that hold in (N, *). Does S=T? If not, is one properly included in the other, and if so, which is included in which? First, some background info. An equational identity, in our context, is one in which only variables appear. The important thing is that we can't mix up the operations, nor can we bring in constants. Also, (x+y)=(y+x) counts as the same identity as (x*y)=(y*x), since switching addition with multiplication results in one transforming into the other. At the very least, I would like a text in which this line of thought is mentioned.

Answer 1

By commutativity and associativity of addition, any expression in an "equational identity" on $(\mathbf{N},+)$ with variables $x_1$, $x_2$, $\dots$, $x_n$ (with repetition) can be re-written in the "standard form" $$(x_1+x_1+\cdots+x_1)+(x_2+x_2+\cdots+x_2)+\cdots+(x_n+x_n+\cdots+x_n)$$ where each grouping contains all copies of the corresponding variable, $x_i$.

Clearly, two expressions are equal if and only if they have the same "standard form"; therefore, an equational identity on $(\mathbf{N},+)$ is characterized by its two expressions having the same variables, with the same repetitions. The same is true for $(\mathbf{N},\times)$.

Answer 2

Consider the cancellation properties of $+$ and $\cdot$ operators on $\mathbb{N}$

If $x+y=z+y$, then we will always have $x=z$.

If, however, $x\cdot y = z\cdot y$, then we may not always have $x=z$. (Assuming $0\in \mathbb{N}$)

So, even in isolation from one another, the $+$ and $\cdot$ operators are not algebraically indistinguishable in the sense that I think you mean.