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How do you describe an operation like this?

$$ a \circ (a \circ b) = b $$

For example, XOR is like this:

$$ a \oplus a \oplus b = b $$

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    $\begingroup$ It might be more convenient to write $a\circ b$ instead of $f(a,\,b)$. (Also, the second equation is redundant.) As for the question: In XOR's case this result follows from the operator being associative and self-inverse and having an identity element. I'm not aware of any name for the final result, which is a shame because you might care about functions that get it for other reasons. $\endgroup$ – J.G. Aug 7 '19 at 20:13
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    $\begingroup$ Is the property you have in mind that $$f(i, f(i, x)) = x$$ holds for some fixed $i$ and all $x$ in the domain $X$ of $f$? If so, the property is just that $f(i, \,\cdot\,)$ is an involution. en.wikipedia.org/wiki/Involution_(mathematics) If the requirement is that it holds for all $i$ and all $x$ in $X$, then one could say that $f(i, \,\cdot\,)$ is an involution for all $i$, or that the image of the curried function $X \to (X \to X)$, $i \mapsto f(i, \,\cdot\,)$, is contained in the set of involutions of $X$---but I don't know a single term to describe this property. $\endgroup$ – Travis Willse Aug 7 '19 at 20:14
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    $\begingroup$ Informally, if $f : X \times X \to X$ satisfies this for all $i,x \in X$ then I might view $f$ as being "an involutive action of $X$ on itself". Slightly more generally, if $f : Y \times X \to X$ satisfies the identity I might say "$f$ is an involutive action of $Y$ on $X$". $\endgroup$ – Daniel Schepler Aug 7 '19 at 20:14
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    $\begingroup$ Do you want $f$ to also have this property in the first argument? i.e., $f(f(i,x),x) = i$? $\endgroup$ – eyeballfrog Aug 7 '19 at 20:16
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    $\begingroup$ The new notation either implies associativity or is ambiguous. I suggest using parentheses, i.e., $a \circ (a \circ b) = b$ to make the intent clear. $\endgroup$ – Travis Willse Aug 7 '19 at 20:33
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In "A guide to self-distributive quasigroups, or latin quandles," David Stanovský calls such an operator left involutory.

A binary algebra $(A,*)$ is called left involutory (or left symmetric) if $x * (x * y) = y$ (hence we have unique left division with $x \backslash y = x ∗ y$).

It's not a very popular term—it currently only has 8 hits on Google—but it makes sense (since it means that the section function $(x * {})$ is involutory), and there are a couple of other papers which use the same terminology with the same meaning (and at least one of these papers cites Stanovský's paper).

So if you're going to write a paper using this term, I suggest using the term "left involutory," and consider citing Stanovský's paper when you do.

(Fun fact: one of the Google search results for "left involutory" caught my eye because it's a chat log from an IRC chat room that I speak in regularly! I couldn't resist mentioning this.)

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If you can rearrange like this, $a \circ (a \circ b) = (a \circ a) \circ b$, then the expression you have implies that $a$ is both the left-inverse and the right-inverse of itself. Hence, $a$ is the inverse of itself.

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    $\begingroup$ If we define an operation on any set $X$ by $x \circ y := y$ then we have $a \circ (a \circ b) = a \circ b = b$, so every element is a left identity but (if $X$ has at least two elements) no right identity, and so no two-sided identity. Since $a \circ (b \circ a) = a$ and $b \circ (a \circ b) = b$, every element is an inverse of every other element (where we mean inverse in the sense of regarding $(S, \circ)$ as a regular semigroup). $\endgroup$ – Travis Willse Aug 8 '19 at 0:48

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