description/proof of that intersection of set minus set and set is intersection of 1st set and 3rd set minus intersection of 2nd set and 3rd set
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of set.
Target Context
- The reader will have a description and a proof of the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
//
Statements:
\((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\)
//
2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\).
3: Proof
Whole Strategy: Step 1: see that \((S_1 \setminus S_2) \cap S_3 \subseteq (S_1 \cap S_3) \setminus (S_2 \cap S_3)\); Step 2: see that \((S_1 \cap S_3) \setminus (S_2 \cap S_3) \subseteq (S_1 \setminus S_2) \cap S_3\).
Step 1:
For any \(p \in (S_1 \setminus S_2) \cap S_3\), \(p \in S_1\) and \(p \in S_3\), so, \(p \in S_1 \cap S_3\), \(p \notin S_2\), \(p \notin S_2 \cap S_3\), so, \(p \in (S_1 \cap S_3) \setminus (S_2 \cap S_3)\).
Step 2:
For any \(p \in (S_1 \cap S_3) \setminus (S_2 \cap S_3)\), \(p \in S_1\), \(p \notin S_2 \cap S_3\), as \(p \in S_3\), \(p \notin S_2\), \(p \in S_1 \setminus S_2\), so, \(p \in (S_1 \setminus S_2) \cap S_3\).
4: Note
\((S_1 \setminus S_2) \cup S_3 = (S_1 \cup S_3) \setminus (S_2 \cup S_3)\) does not necessarily hold although \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\) holds, as is proved in another article.
