Set of generating equations for subtraction.

Question

Consider the algebraic structure $(\mathbb{R}, -)$. Is there a finite generating set of equations for subtraction, and if so, can someone exhibit a list of equations.

Answer 1

I will give a finite list of identities for this structure.

I will write $\langle A; +, -, 0\rangle$ to mean an abelian group with binary addition $+$, unary negation $-$, and (constant) unary zero $0(x)$. I am making $0$ a constant unary operation in order to avoid discussing nullary operations in this answer.

Before getting to the answer, let me point out that the equational theory $\mathcal R$ of the abelian group $\langle \mathbb R; +, -, 0(x)\rangle$ is the same as the equational theory $\mathcal A$ of all abelian groups, and that theory is axiomatized by the finite set:

(Associative law) $x+(y+z)=(x+y)+z$.

(Commutative law) $x+y=y+x$.

($0(x)$ is constant) $0(x)=0(y)$.

($0$ is an additive identity element) $x+0(x)=x$.

(Inverse law) $x+(-x)=0(x)$.

To see why, observe that ${\mathcal A}\subseteq {\mathcal R}$ because $\langle \mathbb R; +, -, 0\rangle$ is an abelian group. On the other hand, if $\varepsilon\notin {\mathcal A}$, then $\varepsilon$ fails in some abelian group, hence $\varepsilon$ fails in some finitely generated free abelian group, hence $\varepsilon$ fails in $\langle \mathbb R; +, -, 0\rangle$, since $\langle \mathbb R; +, -, 0\rangle$ has free abelian subgroups of all finite ranks.

---

For any abelian group $\langle A; +, -, 0\rangle$ there is an associated 'subtraction structure' $\langle A; s(x,y)\rangle$ where $s(x,y)=x+(-y)$. The question asks whether there is a finite basis for the equational theory of the subtraction structure $\langle \mathbb R; s(x,y)\rangle$ associated to $\langle \mathbb R; +, -, 0(x)\rangle$.

There is a dictionary for converting between the abelian group language and the subtraction language. Given an abelian group $\langle A; +, -, 0\rangle$, the associated subtraction structure is $\langle A; s_{(+,-,0)}(x,y)\rangle$ where $s_{(+,-,0)}(x,y):= x+(-y)$. Conversely, given a subtraction structure $\langle A; s(x,y)\rangle$, the associated abelian group is $\langle A; +_s, -_s, 0_s\rangle$ where $0_s(x):=s(x,x)$, $-_s x:=s(0_s,x)=s(s(x,x),x)$, and $x+_s y := s(x,-_s y)=s(x,s(s(x,x),y))$.

Now I will use this dictionary and the fact that $\langle \mathbb R; +, -, 0(x)\rangle$ is a finitely based abelian group to derive that $\langle \mathbb R; s(x,y)\rangle$ is finitely based as a subtraction structure. I will start with $\langle \mathbb R; s(x,y)\rangle$, copy the above finite basis for the associated abelian group $\langle \mathbb R; +_s, -_s, 0_s(x)\rangle$ in the $\{s\}$-language using the given dictionary, then add another identity to say that the original operation $s(x,y)$ is the subtraction associated to the abelian group $\langle \mathbb R; +_s, -_s, 0_s(x)\rangle$. Result:

(Associative law for $+_s$)
$s(x,s(s(x,x),s(y,s(s(x,x),z))))= s(s(x,s(s(x,x),y)),s(s(x,x),z))$.

(Commutative law for $+_s$)
$s(x,s(s(x,x),y))=s(y,s(s(y,y),x))$.

$(0_s(x)$ is constant)
$s(x,x)=s(y,y)$.

($0(x)$ is an additive identity element)
$s(x,s(x,x))=x$.

(Inverse law for $-_s$)
$s(x,s(s(x,x),s(s(x,x),x))))=s(x,x)$.

($s_{(+_s,-_s,0_s)}(x,y)=s(x,y)$)
$s(x,s(s(x,x),s(s(x,x),y))))=s(x,y)$.