Demonstrate $A\subseteq B \Longrightarrow A\circ \theta \subseteq B \circ \theta $

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3 relations: $A, B \subseteq X \times Y$ and $\theta \subseteq Y \times Z$

Demonstrate $A\subseteq B \Longrightarrow A\circ \theta \subseteq B \circ \theta $

I'm having a hard time solving this. $$A\in \langle x,y\rangle $$ $$B\in \langle x,y\rangle $$ $A\in \langle x,y\rangle $ and $ \theta \in \langle y,z\rangle $, so $$ A\circ \theta \in \langle x,z\rangle$$ $B\in \langle x,y\rangle$ and $ \theta \in \langle y,z\rangle$, so $$ B\circ \theta \in \langle x,z\rangle$$

So $A\circ \theta \subseteq B \circ \theta $ is true, so the proposition is true? This is my take on the problem, but if someone can show me how to answer this problem or help me solve it... you will have my thank!

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Your proof could use some more organization/reordering and justification. Here's a cleaned up version.

Suppose that $A \subseteq B$. We want to show that $A \circ \theta \subseteq B \circ \theta$. To this end, choose any $\langle x, z \rangle \in A \circ \theta$. We want to show that $\langle x, z \rangle \in B \circ \theta$.

Indeed, since $\langle x, z \rangle \in A \circ \theta$, we know that there is some $y \in Y$ such that $\langle x, y \rangle \in A$ and $\langle y, z \rangle \in \theta$. But since $A \subseteq B$, we know that $\langle x, y \rangle \in B$. But then since $\langle y, z \rangle \in \theta$ and $y \in Y$, it follows that $\langle x, z \rangle \in B \circ \theta$, as desired. $~~\blacksquare$