Prove the validity of the following expressions: (**a**) $A\circ B$ is a subset (subimage) of $A$. (**b**) If $C$ is a subset of $D$, the $C\circ B$ is a subset of $D\circ B$. (**c**) $(A\circ B)\circ B = A\circ B$.

(**a**) To prove that $A\circ B$ is a subset of $A$, we need to show that every element in $A\circ B$ is also in $A$.

Let $x\in A\circ B$. Then, by definition of composition, there exists $y\in B$ such that $(x,y)\in A$. Since $A$ is a relation from $X$ to $Y$, $x\in X$ and $(x,y)\in A$ implies that $x\in A$. Therefore, $A\circ B\subseteq A$, and (**a**) is proved.

(**b**) To prove that $C\circ B$ is a subset of $D\circ B$, we need to show that every element in $C\circ B$ is also in $D\circ B$.

Let $(x,z)\in C\circ B$. Then, by definition of composition, there exists $y\in B$ such that $(x,y)\in C$ and $(y,z)\in B$. Since $C$ is a subset of $D$, $(x,y)\in C$ implies that $(x,y)\in D$, and therefore $(x,z)\in D\circ B$. Therefore, $C\circ B\subseteq D\circ B$, and (**b**) is proved.

(**c**) To prove that $(A\circ B)\circ B = A\circ B$, we need to show that every element in $(A\circ B)\circ B$ is also in $A\circ B$, and vice versa.

Let $(x,z)\in (A\circ B)\circ B$. Then, by definition of composition, there exists $y\in B$ such that $(x,y)\in A\circ B$ and $(y,z)\in B$. Since $(x,y)\in A\circ B$, there exists $w\in B$ such that $(x,w)\in A$ and $(w,y)\in B$. Then, by transitivity of relations, $(x,z)\in A\circ B$.

Conversely, let $(x,z)\in A\circ B$. Then, there exists $y\in B$ such that $(x,y)\in A$ and $(y,z)\in B$. Since $(x,y)\in A$, we have $(x,z)\in (A\circ B)\circ B$.

Therefore, we have shown that $(A\circ B)\circ B = A\circ B$, and (**c**) is proved.