435: Subset Minus Subset Is Complement of 2nd Subset Minus Complement of 1st Subset

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that subset minus subset is complement of 2nd subset minus complement of 1st subset

---

Topics

About: set

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of set.

---

Target Context

• The reader will have a description and a proof of the proposition that for any set, any subset (the 1st subset) minus any subset (the 2nd subset) is the complement of the 2nd subset minus the complement of the 1st subset.

---

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: Description

For any set, $S$, and any subsets, $S_1,S_2\subseteq S$, $S_1\setminus S_2=(S\setminus S_2)\setminus (S\setminus S_1)$.

---

2: Proof

For any element, $p\in S_1\setminus S_2$, $p\in S_1$ and $p\notin S_2$, $p\in S\setminus S_2$ and $p\notin S\setminus S_1$, so, $p\in (S\setminus S_2)\setminus (S\setminus S_1)$.

For any element, $p\in (S\setminus S_2)\setminus (S\setminus S_1)$, $p\in S\setminus S_2$ and $p\notin S\setminus S_1$, $p\in S_1$ and $p\notin S_2$, so, $p\in S_1\setminus S_2$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>