Prove that if $ \ A\cup B \subseteq C \cup D,\ A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D.$ Question:
Prove that if $ \ A\cup B \subseteq C \cup D,\  A \cap B =$ ∅ $\land \ C \subseteq A \implies B \subseteq D$.
My attempt:
Let $ \ x\in B \implies x \in A \cup B \implies x \in C \cup D \because A\cup B \subseteq C \cup D$.
Now, $ x \in C \lor x\in D$. If $\  x \in C \implies x \in  A \because C \subseteq A$. But that's not possible $\because x \notin A \cap B$, in particular $ x \notin A$. So we must have $ x \in D$.
I found this proof a little challenging. Not quite sure if this is the correct way to prove it. Is my logic correct?


Theorem. Let $A,~B,~C,~D$ be (any) sets. If $A\cup B\subseteq C\cup D$ and $A\cap B=\emptyset$ and $C\subseteq A$, then $B\subseteq D$.

Proof.(in much detailed, giving you the instruction that what should you do in every step)
Let $A,~B,~C,~D$ be sets. Claim:
$A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A\to B\subseteq D$.
Suppose $A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A$. Claim: $B\subseteq D$, namely $\forall x,x\in B\to x\in D$.
Let $x$ in the domain of discourse, claim: $x\in B\to x\in D$.
Suppose $x\in B$(claim $x\in D$). Then $x\in A\vee x\in B$. Hence $x\in A\cup B$. Since $A\cup B\subseteq C\cup D$, we have $x\in C\cup D$, namely $x\in C\vee x\in D$.


*

*If $x\in D$, then done.

*Suppose $x\in C$. by the hypothesis that $C\subseteq A$, we have $x\in A$. Hence $x\in B\wedge x\in A$, namely $x\in A\cap B$. However, by the hypothesis, $A\cap B=\emptyset$. So it is a contradiction(Reductio Ad Absurdum). It is not the case that $x\in C$.


(The following are the ending steps, which is usually omitted in mathematical literature.)
Hence we have shown that $x\in B\to x\in D$. Since $x$ is arbitrary, we have shown that $\forall x,x\in B\to x\in D$, which is $B\subseteq D$.
Hence we have shown that $A\cup B\subseteq C\cup D\wedge A\cap B=\emptyset\wedge C\subseteq A\to B\subseteq D$. Since the sets $A,B,C,D$ are all arbitrary, we have proved the theorem.
