Identities of nested set algebraic expressions

by miraiw
Tags: algebraic, expressions, identities, nested
miraiw is offline
Apr29-12, 06:04 PM
P: 10
Are there any useful identities for simplifying an expression of the form:
$$((\ldots((x_{1} *_{1} x_{2}) *_{2} x_{3}) \ldots) *_{n - 1} x_{n})$$
Where each $$*_{i}$$ is one of $$\cap, \cup$$ and $$x_1 \ldots x_n$$ are sets?

I believe I found two; though I haven't proved them, I think they make sense:
$$((\ldots((x_{1} \cup x_{2}) *_{2} x_{3}) \ldots) \cup x_{1}) \equiv ((\ldots(x_{2} *_{2} x_{3}) \ldots) \cup x_{1})$$
$$((\ldots((x_{1} \cap x_{2}) *_{2} x_{3}) \ldots) \cap x_{1}) \equiv ((\ldots(x_{2} *_{2} x_{3}) \ldots) \cap x_{1})$$

Generally, how would you prove these? Just induction?

I tested a few expressions with the attached perl script which is why I think they work.
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Stephen Tashi
Stephen Tashi is offline
Apr30-12, 08:16 PM
Sci Advisor
P: 3,177
There are various distributive laws like
[tex] (A \cup B) \cap C = (A\cap C) \cup (B \cap C) [/tex]
[tex] (A \cap B) \cup C = (A \cup C) \cap (B \cup C) [/tex]

Since you are using [itex] *_i [/itex] to represent something that can be either [itex] \cup [/itex] or [itex] \cap [/itex] , I can't tell offhand if your identities are correct. I'd have to consider each possible interpretation of the operation as a separate case.

Induction would be the method of proof, assuming that your "..." indicates a an expression of arbitrary but finite length. If you have in mind some sort of infinitely long expressions, you have to start with the problem of defining what they would mean.
miraiw is offline
Apr30-12, 10:54 PM
P: 10
I restated the problem in terms of propositions like $$x \in A_{1} \wedge x \in A_{2}$$ and considered that whatever was in the ellipsis would either depend on the innermost expression or not and if it did would either have a value opposite the innermost or the same as. From there I just exhaustively listed the cases and compared. Also, I assumed that the [itex] A_{1}[/itex] doesn't show anywhere in the ellipsis.

I think that's all I need since I can just deal with the rest recursively.

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