242: Product of Any Complements Is Product of Whole Sets Minus Union of Products of Whole Sets 1 of Which Is Replaced with Subset for Each Product

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A description/proof of that product of any complements is product of whole sets minus union of products of whole sets 1 of which is replaced with subset for each product

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a description and a proof of the proposition that the product of any complements is the product of the whole sets minus the union of the products of the whole sets, 1 of which is replaced with a subset for each product.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any sets, $S_1,S_2,...,S_n$, and any subsets, $S_i^{\prime }\subseteq S_i$, $(S_1\setminus S_1^{\prime })\times (S_2\setminus S_2^{\prime })\times ...\times (S_n\setminus S_n^{\prime })=S_1\times S_2\times ...\times S_n\setminus ((S_1^{\prime }\times S_2\times ...\times S_n)\cup (S_1\times S_2^{\prime }\times ...\times S_n)\cup ...\cup (S_1\times S_2\times ...\times S_n^{\prime }))$.

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2: Proof

For any $p=(p_1,p_2,...,p_n)\in (S_1\setminus S_1^{\prime })\times (S_2\setminus S_2^{\prime })\times ...\times (S_n\setminus S_n^{\prime })$, $p_i\in S_i\setminus S_i^{\prime }$, $p_i\notin S_i^{\prime }$, $(p_1,p_2,...,p_n)\notin S_1\times S_2\times ...\times S_i^{\prime }\times ...\times S_n$, so, $p\in S_1\times S_2\times ...\times S_n\setminus ((S_1^{\prime }\times S_2\times ...\times S_n)\cup (S_1\times S_2^{\prime }\times ...\times S_n)\cup ...\cup (S_1\times S_2\times ...\times S_n^{\prime }))$; for any $p\in S_1\times S_2\times ...\times S_n\setminus ((S_1^{\prime }\times S_2\times ...\times S_n)\cup (S_1\times S_2^{\prime }\times ...\times S_n)\cup ...\cup (S_1\times S_2\times ...\times S_n^{\prime }))$, $p\notin (S_1^{\prime }\times S_2\times ...\times S_n)\cup (S_1\times S_2^{\prime }\times ...\times S_n)\cup ...\cup (S_1\times S_2\times ...\times S_n^{\prime })$, $p\notin S_1\times S_2\times ...\times S_i^{\prime }\times ...\times S_n$ for each $i$, $p_i\notin S_i^{\prime }$ for each $i$, $p_i\in S_i\setminus S_i^{\prime }$ for each $i$, so, $p\in (S_1\setminus S_1^{\prime })\times (S_2\setminus S_2^{\prime })\times ...\times (S_n\setminus S_n^{\prime })$.

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References

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