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In relational algebra, for any pair of relations $R_{1}$ and $R_{2}$, the standard division operation is denoted by $\div$ and defined as follows: $$R_{1} \div R_{2}=\pi_{A^{\ast }}\left(R_{1}\right)-\pi_{A^{\ast}}\left(\left(\pi_{A^{\ast }}\left(R_{1}\right) \times R_{2}\right)-R_{1}\right)$$ Here, $A^{\ast }$ denotes the attributes that are in $R_{1}$ but not in $R_{2}$. The following is an example of a standard division operation.

$R_{1}$
A B C
1 Red Leaf
2 Green Leaf
2 Blue Leaf
3 Red Leaf
3 Red Stem
4 Pink Leaf
5 Black Stem
$R_{2}$
C
Leaf
Stem
$R_{1} \div R_{2}$
A B
3 Red
1. Cite a working example wherein a pair of relations $R_{1}$ and $R_{2}$ satisfies the following equation: $$\left(R_{1} \div R_{2}\right) \times R_{2}=R_{1}$$
2. Recall that the notations $\Join, ⟕$  and $⟖$ denote natural join, left outer join, and right outer join operations, respectively. For any arbitrary pair of relations $R_{1}$ and $R_{2}$, prove that if $\left(R_{1} \div R_{2}\right) \times R_{2}=R_{1}$ holds, then the following will also hold. $$\left(R_{1} ⟕ R_{2}\right)-\left(R_{1} \Join R_{2}\right) \equiv\left(R_{1} ⟖ R_{2}\right)-\left(R_{1} \Join R_{2}\right) .$$