919: Set Minus (Set Minus Set) Is Not Necessarily but Contains (1st Set Minus 2nd Set) Minus 3rd Set

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description/proof of that set minus (set minus set) is not necessarily but contains (1st set minus 2nd set) minus 3rd 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: 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 any set minus (any set minus any set) is not necessarily equal to but contains (the 1st set minus the 2nd set) minus the 3rd 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:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$S_3$: $\in \{\text{ the sets }\}$
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Statements:
not necessarily $S_1\setminus (S_2\setminus S_3)=(S_1\setminus S_2)\setminus S_3$
$\wedge$
$(S_1\setminus S_2)\setminus S_3\subseteq S_1\setminus (S_2\setminus S_3)$
//

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2: Natural Language Description

For any sets, $S_1,S_2,S_3$, $S_1\setminus (S_2\setminus S_3)$ is not necessarily $(S_1\setminus S_2)\setminus S_3$, but $(S_1\setminus S_2)\setminus S_3\subseteq S_1\setminus (S_2\setminus S_3)$.

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3: Proof

Whole Strategy: Step 1: see an example that $S_1\setminus (S_2\setminus S_3)\ne (S_1\setminus S_2)\setminus S_3$; Step 2: see that $(S_1\setminus S_2)\setminus S_3\subseteq S_1\setminus (S_2\setminus S_3)$.

Step 1:

For the 1st part, a counterexample suffices.

Let $S_1=\{0,1\},S_2=\mathrm{\varnothing },S_3=\{0\}$. Then, $S_1\setminus (S_2\setminus S_3)=\{0,1\}\setminus \mathrm{\varnothing }=\{0,1\}$; $(S_1\setminus S_2)\setminus S_3=\{0,1\}\setminus \{0\}=\{1\}$.

Step 2:

For the 2nd part, for any $p\in (S_1\setminus S_2)\setminus S_3$, $p\in S_1$, $p\notin S_2$, so, $p\notin S_2\setminus S_3$, so, $p\in S_1\setminus (S_2\setminus S_3)$.

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References

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