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Notation: For any two numbers $a$ and $b$, let the maximum be $a\sqcap b$, and let the minimum be $a\sqcup b$. (No, my symbols aren't upside-down. Compare this with floor notation; $\lfloor a\rfloor\leq a$, and likewise $a\sqcup b\leq a$.)

Addition distributes over max and min, and max and min distribute over each other:

$$ c + (a \sqcap b) = (c + a) \sqcap (c + b) \qquad -(a \sqcap b) = (-a) \sqcup (-b) $$ $$ c + (a \sqcup b) = (c + a) \sqcup (c + b) \qquad -(a \sqcup b) = (-a) \sqcap (-b) $$ $$ c \sqcap (a \sqcup b) = (c \sqcap a) \sqcup (c \sqcap b) \qquad c \sqcup (a \sqcap b) = (c \sqcup a) \sqcap (c \sqcup b) $$

It follows that any expression involving these operations can be put in a form where the "$+$" and "$-$" signs are innermost, and only "$\sqcap$" and "$\sqcup$" apply to parentheses. So, to reduce the number of parentheses, let's extend the standard order of operations: Multiplication and division are done first (left to right), then addition and subtraction (left to right), then maximum and minimum (left to right).

$$ (a + b) \sqcap (c + d) = a + b \sqcap c + d \neq a + (b \sqcap c) + d $$ $$ (a \cdot b) \sqcap (c \cdot d) = a \cdot b \sqcap c \cdot d \neq a \cdot (b \sqcap c) \cdot d $$

However, multiplication doesn't properly distribute:

$$ c\cdot(a \sqcap b) = \begin{cases} c\cdot a \sqcap c\cdot b, \quad c\geq0 \\ c\cdot a \sqcup c\cdot b, \quad c\leq0 \end{cases} $$

Question: Is $c\cdot(a \sqcap b)$ represented by an expression involving the symbols "$a,b,c,+,-,\cdot,\sqcap,\sqcup,(,),0,1$" (and maybe other real constants), where "$+$" and "$-$" and "$\cdot$" are never adjacent to "$($" or "$)$"?

(Don't cheat and try something like "$c\cdot1(a \sqcap b)$", which should be "$c\cdot1\cdot(a \sqcap b)$".)

Equivalently, are there polynomials $p_{i,j}$ in three variables such that

$$ \begin{align} c\cdot(a \sqcap b) & = \big(p_{1,1}(a,b,c) \sqcap p_{1,2}(a,b,c) \sqcap p_{1,3}(a,b,c) \sqcap \cdots\big) \\ &\; \sqcup \big(p_{2,1}(a,b,c) \sqcap p_{2,2}(a,b,c) \sqcap p_{2,3}(a,b,c) \sqcap \cdots\big) \\ &\; \sqcup \big(p_{3,1}(a,b,c) \sqcap p_{3,2}(a,b,c) \sqcap p_{3,3}(a,b,c) \sqcap \cdots\big) \\ &\; \sqcup \;\vdots \end{align} $$

for all $a,b,c\in\mathbb R$ ?

(I think it would follow that any expression, not just $c\cdot(a \sqcap b)$, can be put in the same form, by repeatedly applying these distributive laws. But I'm not really sure. The process might never end, always making the expression more complicated.)

We can decompose any number as the difference of two non-negative numbers:

$$ c = (0 \sqcap c) + (0 \sqcup c) = (0 \sqcap c) - (0 \sqcap -c) $$

Thus, since positive numbers distribute over max and min, we have

$$ c\cdot(a \sqcap b) = (0 \sqcap c)\cdot(a \sqcap b) - (0 \sqcap -c)\cdot(a \sqcap b) $$ $$ = ((0 \sqcap c)\cdot a \sqcap (0 \sqcap c)\cdot b) - ((0 \sqcap -c)\cdot a \sqcap (0 \sqcap -c)\cdot b) $$ $$ = (a\cdot(0 \sqcap c) \sqcap b\cdot(0 \sqcap c)) - (a\cdot(0 \sqcap -c) \sqcap b\cdot(0 \sqcap -c)) $$

This shows that it suffices to find a formula for $a\cdot(0 \sqcap c)$, or equivalently $a\cdot(0 \sqcap b)$; the number of variables is reduced from three to two. In fact there's another way to show this:

$$ a \sqcap b = (a-b \sqcap 0) + b $$ $$ c\cdot(a \sqcap b) = c\cdot(0 \sqcap a-b) + c\cdot b $$

If that term $c\cdot(0 \sqcap a-b)$ can be expressed in the desired form, then (since addition distributes) we can simply add $c\cdot b$ to all the polynomials $p_{i,j}$.

Furthermore, this is also equivalent to finding a formula for $(0 \sqcap a)\cdot(0 \sqcap b)$:

$$ a\cdot(0 \sqcap b) = (0 \sqcap a)\cdot(0 \sqcap b) - (0 \sqcap -a)\cdot(0 \sqcap b) $$

Related: Are there any theories being developed which study structures with many operations and many distributive laws?

I couldn't find good tags for this.

mr_e_man
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1 Answers1

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Yes. But the expressions I found are somewhat complicated and asymmetric. If anyone can find anything nicer, please post another answer.

$$ (0 \sqcap a)\cdot(0 \sqcap b) = \begin{cases} a\cdot b, \quad a\geq0 \;\text{and}\; b\geq0 \\ \;\;0,\;\; \quad a\leq0 \;\;\text{or}\;\; b\leq0 \end{cases} $$ $$ = (0 \sqcap a\cdot b) \sqcup (0 \sqcap a \sqcap a\cdot b\cdot b) $$

Suppose $a$ and $b$ are positive. Then $ab\leq a$ when $b\leq1$, and $ab\leq ab^2$ when $b\geq1$, so in either case $ab\leq a \sqcap ab^2$. This is the basis of most of these formulas.

$$ a\cdot(0 \sqcap b) = \begin{cases} a\cdot b, \quad b\geq0 \\ \;\;0,\;\; \quad b\leq0 \end{cases} $$ $$ = \big(ab \sqcup (0 \sqcap a \sqcap ab^2)\big) \sqcap (0 \sqcup a \sqcup ab^2) $$ $$ = \big(ab \sqcap (0 \sqcup a \sqcup ab^2)\big) \sqcup (0 \sqcap a \sqcap ab^2) $$ $$ = \big(ab \sqcup (0 \sqcap b \sqcap a^2b)\big) \sqcap (0 \sqcup -b \sqcup -a^2b) $$ $$ = \big(ab \sqcap (0 \sqcup -b \sqcup -a^2b)\big) \sqcup (0 \sqcap b \sqcap a^2b) $$ $$ = (0 \sqcap ab) \sqcup (ab \sqcap a) \sqcup (ab \sqcap ab^2) \sqcup (0 \sqcap a \sqcap ab^2) $$

Replacing $a$ with $c$ and $b$ with $a-b$:

$$ c\cdot(0 \sqcap a-b) $$ $$ = \big(0 \sqcap c(a-b)\big) \sqcup \big(c(a-b) \sqcap c\big) \sqcup \big(c(a-b) \sqcap c(a-b)^2\big) \sqcup \big(0 \sqcap c \sqcap c(a-b)^2\big) $$

Adding $c\cdot b$:

$$ c\cdot(a \sqcap b) $$ $$ = \big(cb \sqcap ca\big) \sqcup \big(ca \sqcap c+cb\big) \sqcup \big(ca \sqcap c(a-b)^2+cb\big) \sqcup \big(cb \sqcap c+cb \sqcap c(a-b)^2+cb\big) $$

mr_e_man
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