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With respect to an operation $O_n$ on an infinite set, define a mimic operation $O'_n$, as an operation which differs from $O_n$ only by a finite number of points, and a $k$-mimic operation $O^{k}_n$ of $O_n$ as an operation which differs from $O_n$ by $k$ points. Equivalently, for a $k$-mimic, for the set of tuples which completely describe $O_n$, $\{(a_1,\dots,a_{n+1}),\dots,(z_1,\dots, z_{n+1})\}$, there exist $k$ tuples such that $O^{k}_n(x_1,\dots,x_n)\ne O_n(x_1,\dots,x_n)$, otherwise $O^{k}_n(x_1,\dots,x_n)=O_n(x_1,\dots,x_n)$.

It doesn't seem too hard to prove that given an idempotent operation $A$, there exist an infinity of $k$-mimics $B$, such that each operation $B$ is an idempotent mimic of $A$ (the mimic operation also satisfies idempotency). We can also prove that if an operation $O_n$ commutes, and if (x$O_n$x)=(x$O_n$$^{1}$x), then $O_n$$^{1}$ does not commute. This seems to suggest that for a commutative operation only even numbered commutative $k$-mimics exist, when (x$O_n$x)=(x$O_n$$^{1}$x). I think that $\min(x, y$) and $$G(x, y)=\begin{cases}1,&\text{if }x=y=0\\\min(x, y),&\text{otherwise}\end{cases}$$ where $x,y \in\mathbb R$, are associative $k$-mimics, which suggests that for an associative operation on an infinite set, an infinity of $1$-mimics exist (please correct me if this is wrong, I think I proved this the other day, but I'm not quite sure my method worked out.) How many associative $k$-mimics of an associative operation exist?

Does there exist another term for what I've called a "mimic operation"?

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Sorry, I had little idea what to tag this. –  Doug Spoonwood Apr 14 '12 at 3:16
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Why can't you just change a single value $F(x,x)$ to get a commutative $1$-mimic of a commutative operation? In particular, your own $G$ seems to be a commutative $1$-mimic of $\min$? –  joriki Apr 14 '12 at 4:42
    
@joriki You can do that, I totally missed that, great comment! Alright, so that means that all commutative 1-mimics of a commutative operation happen "along the diagonal" or differ only at F(x, x). –  Doug Spoonwood Apr 14 '12 at 5:18
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