Distributive property for signed multiset I'm reading through Wayne Blizard's introduction to signed multisets, which says at the bottom of page 9 that additive union distributes over intersection. I see how this is true for normal (non-negative multiplicity) multisets, but for signed multisets I get a contradiction. What would I need to do differently to restore the distributive property?
The way I'm calculating is:
Additive union: The multiplicity of an object in the returned multiset is the sum of its multiplicities in the argument multisets.
Intersection: If an object has nonzero multiplicity in both argument multisets, its multiplicity in the returned multiset is the minimum of its multiplicities in the argument multisets; otherwise, its multiplicity in the returned multiset is zero.
$$ [z]_{1} \uplus ([z]_{0} \cap [z]_{-1}) =  [z]_1 \uplus [z]_{0} = [z]_{1} $$
$$ [z]_{1} \uplus ([z]_{0} \cap [z]_{-1}) = ([z]_1 \uplus [z]_{0}) \cap ([z]_1 \uplus [z]_{-1}) = [z]_1 \cap [z]_{0} = [z]_{0} $$
 A: As you've proven, your current definition of intersection does not obey the distributive property. The reason for this is basically because the multiplicity zero is being singled out as exceptional, but if we remove the exception, the distributive property is restored. Redefine intersection so that $[z]_a\cap [z]_b=[z]_{\min(a,b)}$ regardless of whether $a,b$ are zero or nonzero, then we prove the distributive property like so.
$$[z]_a \uplus([z]_b\cap [z]_c) = [z]_a\uplus [z]_{\min(b,c)} = [z]_{a+\min(b+c)} = [z]_{\min(a+b,a+c)}$$
$$ = [z]_{a+b}\cap [z]_{a+c} = ([z]_a\uplus [z]_b)\cap([z]_a\uplus[z]_c)$$
Conversely, this definition for intersection is unique, this is the only way to define intersection in such a way that we get the distributive property. Indeed, suppose addition is defined as you have it, and suppose that the distributive property holds in general, and suppose that intersection is defined as normal at least when multiplicities are positive. Given these assumptions, it necessarily follows that $[z]_a\cap[z]_b=[z]_{\min(a,b)}$ regardless of whether $a,b$ are zero or nonzero. To prove this, select some $c>|a|+|b|$ so that both $a+c$ and $b+c$ are positive, then we exploit the distributive property like so.
$$[z]_a\cap [z]_b = ([z]_{a+c}\uplus[z]_{-c})\cap([z]_{b+c}\uplus[z]_{-c}) = ([z]_{a+c}\cap[z]_{b+c})\uplus[z]_{-c}$$
$$ = [z]_{\min(a+c,b+c)}\uplus[z]_{-c} = [z]_{\min(a+c,b+c)-c} = [z]_{\min(a,b)}$$
The above analysis proves that, if addition is defined as normal, and intersection is defined as normal for positive multiplicities, then the distributive property will hold if and only if we generally have $[z]_a\cap [z]_b=[z]_{\min(a,b)}$. If you define intersection in any other way, either the distributive property will fail, or your definition of intersection will disagree with the usual definition for positive multiplicities.
