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Let $x$ and $y$ denote variables over the set $\{a, b, c\}=S$, and $t$ some constant of $S$. Let "X" denote an unknown binary operation on $S$. Suppose we have the following set of properties:

  1. For all x, xxX=t
  2. There exists a y, such that for all x, xyX=y=t (the same constant as in 1.)
  3. There exists an x, such that for all y, xyX=t
  4. There exists an x, such that for all y, xyX=y
  5. There exists an x, such that there exists a y not equal to x, and xyX=x.

There exist at least 6 structures (all automorphic to each other, which indicates how I obtained them in the first place, THEN I wrote the properties) which satisfy this set of properties as follows:

A  a  b  c a  c  c  c b  b  c  c c  a  b  c  B  a  b  c a  b  b  b b  a  b  c c  c  b  b  C  a  b  c a  c  a  c b  c  c  c c  a  b  c  D  a  b  c a  a  b  c b  a  a  a c  a  c  a  E  a  b  c a  b  b  a b  a  b  c c  b  b  b  F  a  b  c a  a  b  c b  a  a  b c  a  a  a 

Do any other binary operations on {a, b, c} satisfy properties 1.-5. above? I would think "yes", since, if I've gotten things right, property 1. specifies something about the diagonal, property 2. specifies a particular column, for which all of its values equal that of the diagonal, 3. specifies a row that always equals the valued of the diagonal, 4. specifies a row which always equals the values of the second coordinate, and I think 5. specifies the only value left undetermined by 1.-4. Do any other structures exist here?

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    Since someone voted on this, and I looked at this again, I'll add this comment. I don't think I had realized it when I wrote this question, but you can interpret each of these structures as representing the conditional for Lukasiewicz 3-valued logic (also Reichenbach 3-valued logic). So, the above indicates one way to write them in terms of first-order logic. Also, if you look at each of the tables, you can either cover the element which appears most with 1 discrete equilateral triangle, or 1 discrete equilateral triangle and$a$discrete right triangle.2013-07-19

1 Answers 1

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So, suppose you have a 3 element magma $M$, and an element $t$ such that:

  1. For all $x\in M$, $xx=t$.
  2. There exists $y\in M$ such that for all $x\in M$, $xy=t$.
  3. There exists $w\in M$ such that for all $x\in M$, $wx=t$.
  4. There exists $z\in M$ such that for all $x\in M$, $zx=x$.
  5. There exist $x,y\in M$, $x\neq y$, such that $xy=x$.

Note that $z\neq w$: letting $x\neq t$ (possible since $M$ has three elements), then $wx = t \neq x = zx$ by 3 and 4, so $z\neq w$.

Note that $z=t$: for $t=zz=z$, by 1 and 4.

Likewise, $y=t$, since $t = zy = y$ (by 2 and 4). Thus, $y=z=t$, and $w$ is different from $t$.

Thus, $t$ is a right zero and a left identity. So we have:

$M$ is a 3-element magma; there are distinguished elements $t$ and $w$ such that:

  1. $t\neq w$.
  2. $xx = t$ for all $x\in M$.
  3. $xt = t$ for all $x\in M$.
  4. $tx = x$ for all $x\in M$.
  5. $wx = t$ for all $x\in M$.
  6. There exists $z,r\in M$, $z\neq r$, such that $zr=z$.

From 4 and the clause $z\neq r$ in 6, we have that $z\neq t$. From this and 5, we have that $z\neq w$. Thus, the three elements of $M$ are $t$, $w$, and $z$. From 3, we know that $r\neq t$, and since $r\neq z$, hence $r=w$. Thus:

$M$ is a three element magma. The three elements are $t$, $w$, and $z$. They satisfy:

  • $xx = t$ for all $x$.
  • $xt = t$ for all $x$.
  • $tx = x$ for all $x$.
  • $wx = t$ for all $x$.
  • $zw = z$.

Thus, the multiplication table of $M$ is: $\begin{array}{c|ccc} &t&w&z\\ \hline t&t&w&z\\ w&t&t&t\\ z&t&z&t \end{array}$ The first row is forced by the third listed property in the last quote box. The first column is forced the second listed property; the second row by the fourth listed property; the last diagonal entry by the first listed property; and the $zw$ entry by the last listed property. This gives all the entries of the table.

Thus, the properties completely determines $M$ up to the names of the elements. If you want to call the elements $a$, $b$, and $c$, then there are exactly $3!=6$ ways in which you can assign the names $a$, $b$, and $c$ to $t$, $w$, and $z$, so that gives exactly six different tables in terms of $a$, $b$, and $c$, though they are all isomorphic to $M$.

(And it would have taken me about half as long to figure out what you are asking if you were using standard notations and nomneclature instead of insisting on your own personal language; I wonder if your favorite movie is Nell.)