I asked this on Math Stack Exchange, but it didn't get a single answer. So, I am now asking it here. Let our signature be that of a single binary operation $+$. I define the constant identity to be $x+y=z+w$. The commutative identity is, of course, the well-known identity $x+y=y+x$. I wonder, is there an identity strictly between those two, meaning, is there a single identity $E$ such that the constant identity implies $E$, but not conversely, and $E$ implies the commutative identity, but not conversely?
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2$\begingroup$ I'm trying to understand the setup ... what does "define the constant identity to be $x+y=z+w$" mean? $\endgroup$– Greg MartinCommented Jul 18, 2023 at 8:14
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$\begingroup$ Don't you mean "I define the constant identity to be $x+y=x+y$"? That would make more sense to me, although I'd rather call it "the trivial identity". $\endgroup$– md2perpeCommented Jul 18, 2023 at 8:48
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4$\begingroup$ @GregMartin Subject to that identity, the image of the addition function is a single object. So, user107952 has named that identity the "constant identity" to emphasize this fact. The word "define" is a slight misnomer; the identity has just been given a convenient name. $\endgroup$– Pace NielsenCommented Jul 18, 2023 at 11:19
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3$\begingroup$ Oh, so the "constant identity" is just the statement that the binary operation $+$ is a constant function of two variables. I'm surprised that this is an interesting thing to consider, but judging from the votes on the question and accepted answer, people are indeed interested :) $\endgroup$– Greg MartinCommented Jul 18, 2023 at 16:27
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1$\begingroup$ [Just a side remark: there is another stronger constant identity, namely $x=y$.] $\endgroup$– YCorCommented Oct 11 at 16:38
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2 Answers
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Yes: $$(x + x) + y = y + x$$ The constant identity implies this because both sides are $+$es. This does not imply the constant identity because it is true about any set with an operation that is commutative, associative, and idempotent (meaning $x + x = x$ for all $x$), the smallest nontrivial example is $\{x,y\}$ where $x + x = x + y = y + x = x$ and $y + y = y$.
This implies commutativity. $(x + x) + (x + x) = (x + x) + x = x + x$, so $x + x$ is idempotent. $(x+x)+y=((x+x)+(x+x))+y=y+(x+x)$, so $x + x$ commutes with everything. $(x + x) + (y + y) = (y + y) + x = x + y$, and $(x + x) + (y + y) = (y + y) + (x + x)$ because $x+x$ commutes with everything, so $x + y = y + x$, so $+$ is commutative.
This is not implied by commutativity, because (for example) addition of natural numbers is commutative but does not satisfy this identity.
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2$\begingroup$ Cool! For those, like myself, wondering why $x+y=y+z$ wouldn't do, the answer is that $x+y=y+z$ & $y+z=z+w$ imply $x+y=z+w$. $\endgroup$ Commented Jul 17, 2023 at 9:34
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$\begingroup$ For the less advanced among us, what does this identity (which seems false at least with traditional definitions of the symbols!) mean? Is this just defining an algebra (is it an algebra?) where the sum of any two symbols is the same? (e.g., such that all symbols are equal to the natural number 0 and addition is the traditional operation, or all symbols are 1 and + is multiplication)? $\endgroup$ Commented Jul 18, 2023 at 16:08
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2$\begingroup$ @AndrewJaffe Given an algebra $(A,+)$ (i.e., a set $A$ and an arbitrary binary function $+\colon A^2\to A$), one can ask whether or not it satisfies this identity. An example satisfying this identity is given by paste bee at the end of the first paragraph. Note that in this example, it is not the case that the sum of any two elements is the same (i.e., in the language of the OP, this example algebra does not satisfy the constant identity $x+y = z+w$). $\endgroup$ Commented Jul 18, 2023 at 17:03
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$\begingroup$ To add to paste bee's answer: a somewhat natural examples of operation satisfying this idempotence identity are binary min/max functions. min(min(x, x), y) = min(y, x) $\endgroup$ Commented Oct 11 at 16:25
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From the data of the Equational Theories Project, one can give the complete list of such laws of order four or less (i.e., using at most four operations):
The commutative law (43) is the red box on top; the constant law (46) is in the blue box on the bottom (together with several other equivalent laws), and the equations in the intermediate boxes (including paste bee's law (332)) are intermediate. An interactive version of this image can be found here, where one can click through to find more information on each equation, and links to Lean formalizations of various implications (or refutations of implications).