735: For Set and Set, Power Set of [Former Set Minus Latter Set] Is [Power Set of Former Set] Elements Minus Latter Set

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description/proof of that for set and set, power set of [former set minus latter set] is [power set of former set] elements minus latter set

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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: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of power set of set.
• The reader knows a definition of set elements minus set.
• The reader admits the proposition that for any set, the union of the power set of the set is the set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set and any set, the power set of [the former set minus the latter set] is [the power set of the former set] elements minus the latter set.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$S^{\prime }$: $\in \{\text{ the sets }\}$
$S$: $\in \{\text{ the sets }\}$
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Statements:
$Pow(S^{\prime }\setminus S)=Pow(S^{\prime }){\setminus }_{e}S$
//

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2: Natural Language Description

For any set, $S^{\prime }$, and any set, $S$, $Pow(S^{\prime }\setminus S)=Pow(S^{\prime }){\setminus }_{e}S$.

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3: Note

The "[ ]" marks in the title and Target Context are in order for removing the ambiguities: for example, 'Power Set of Former Set Minus Latter Set' is ambiguous whether it is "Power Set of [Former Set Minus Latter Set]" or '[Power Set of Former Set] Minus Latter Set'.

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4: Proof

Whole Strategy: Step 1: see that each element of $Pow(S^{\prime }\setminus S)$ is contained in $Pow(S^{\prime }){\setminus }_{e}S$; Step 2: see that each element of $Pow(S^{\prime }){\setminus }_{e}S$ is contained in $Pow(S^{\prime }\setminus S)$; Step 3: conclude the proposition.

Step 1:

Let $S^{″}\in Pow(S^{\prime }\setminus S)$ be any.

That means that $S^{″}$ is a subset of $S^{\prime }\setminus S$. So, there is a formula, $\varphi$, such that $S^{″}=\{p\in S^{\prime }\setminus S|\varphi (p)\}$.

$Pow(S^{\prime }){\setminus }_{e}S=\{p\in Pow(\cup Pow(S^{\prime }))|\mathrm{\exists }{p}^{\prime }\in Pow(S^{\prime })(p=p^{\prime }\setminus S)\}$, by the definition of set elements minus set. As $\cup Pow(S^{\prime })=S^{\prime }$, by the proposition that for any set, the union of the power set of the set is the set, $Pow(S^{\prime }){\setminus }_{e}S=\{p\in Pow{S}^{\prime }|\mathrm{\exists }{p}^{\prime }\in Pow(S^{\prime })(p=p^{\prime }\setminus S)\}$.

$S^{″}\in Pow{S}^{\prime }$, because $S^{″}=\{p\in S^{\prime }|p\in S^{\prime }\setminus S\wedge \varphi (p)\}$, because for each $p^{\prime }\in S^{″}$, $p^{\prime }\in S^{\prime }$, $p^{\prime }\in S^{\prime }\setminus S$, and $\varphi (p)$; for each $p^{\prime }\in \{p\in S^{\prime }|p\in S^{\prime }\setminus S\wedge \varphi (p)\}$, $p^{\prime }\in S^{\prime }\setminus S$ and $\varphi (p)$, which means that $p^{\prime }\in S^{″}$. $\mathrm{\exists }{p}^{\prime }\in Pow(S^{\prime })(S^{″}=p^{\prime }\setminus S)$, because $p^{\prime }=S^{″}$ will do.

So, $S^{″}\in Pow(S^{\prime }){\setminus }_{e}S$. So, $Pow(S^{\prime }\setminus S)\subseteq Pow(S^{\prime }){\setminus }_{e}S$.

Step 2:

Let $S^{″}\in Pow(S^{\prime }){\setminus }_{e}S$ be any.

Is $S^{″}$ a subset of $S^{\prime }\setminus S$?

There is a $p^{\prime }\in Pow(S^{\prime })$ such that $S^{″}=p^{\prime }\setminus S$. As $p^{\prime }$ is a subset of $S^{\prime }$, there is a formula, $\varphi$, such that $p^{\prime }=\{p\in S^{\prime }|\varphi (p)\}$. As $S^{″}=p^{\prime }\setminus S$, $S^{″}=\{p\in S^{\prime }|\varphi (p)\wedge p\in S^{\prime }\setminus S\}$. But that equals $S^{″}=\{p\in S^{\prime }\setminus S|\varphi (p)\}$, because for each $p^{″}\in \{p\in S^{\prime }|\varphi (p)\wedge p\in S^{\prime }\setminus S\}$, $p^{″}\in S^{\prime }\setminus S$ and $\varphi (p)$; for each $p^{″}\in \{p\in S^{\prime }\setminus S|\varphi (p)\}$, $p^{″}\in S^{\prime }$, $\varphi (p)$, and $p^{″}\in S^{\prime }\setminus S$.

So, $S^{″}$ is a subset of $S^{\prime }\setminus S$.

So, $S^{″}\in Pow(S^{\prime }\setminus S)$. So, $Pow(S^{\prime }){\setminus }_{e}S\subseteq Pow(S^{\prime }\setminus S)$.

Step 3:

So, $Pow(S^{\prime }\setminus S)=Pow(S^{\prime }){\setminus }_{e}S$.

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References

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