Consider the algebraic structure $(\mathbb{R};d,0)$, where $d$ is the distance $|x - y|$ between two real numbers. Is there a finite equational basis for the identities of that structure? I conjecture that the commutative law and the equation $d(x,x)=0$ is sufficient. Is it, and if not, is there a further finite basis, or is there no finite basis?
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What is an "equational basis"? – Lee Mosher Aug 18 '21 at 00:22
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@LeeMosher A set of equations which generate the equational theory of the structure under logical entailment. See universal algebra. – Alex Kruckman Aug 18 '21 at 00:26
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What about $d(0,x)=x$? – Alex Kruckman Aug 18 '21 at 00:28
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3@AlexKruckman: Well, that's only true for nonnegative $x$, but you could say $d(0,d(x,y))=d(x,y)$. – Eric Wofsey Aug 18 '21 at 01:40
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This seems to reduce to the question about finite equational base for the function $abs$ such that $abs(x)=|x|$, and then define $d(x,y)=abs(s(x,y))$, where $s$ is given the finite equational base in this answer (to a question of yours). Can that be done without having the order relation defined? – amrsa Aug 20 '21 at 11:00