Not a duplicate of
Suppose $A$, $B$, and $C$ are sets. Prove that $C ⊆ A △ B$ iff $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$.
This is exercise $3.5.21$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $A$, $B$, and $C$ are sets. Prove that $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.
Here is my proof:
$(\rightarrow)$ Suppose $C\subseteq A\Delta B$.
$(1)$ Let $x$ be an arbitrary element of $C$. From $C\subseteq A\Delta B$ and $x\in C$, $x\in A\Delta B$. Now we consider two cases.
Case $1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\cup B$.
Case $2.$ Suppose $x\in B\setminus A$. Ergo $x\in A\cup B$.
Since the above cases are exhaustive, $x\in A\cup B$. Thus if $x\in C$ then $x\in A\cup B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\cup B)$ and so $C\subseteq A\cup B$. Therefore if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$.
$(2)$ Suppose $A\cap B\cap C\neq\emptyset$. So we can choose some $x_0$ such that $x_0\in A$, $x_0\in B$, and $x_0\in C$. From $C\subseteq A\Delta B$ and $x_0\in C$, $x_0\in A\Delta B$. Now we consider two cases.
Case $1.$ Suppose $x_0\in A\setminus B$. Ergo $x_0\notin B$ which contradicts $x_0\in B$ and so it must be the case that $A\cap B\cap C=\emptyset$.
Case $2.$ Suppose $x_0\in B\setminus A$. Ergo $x_0\notin A$ which contradicts $x_0\in A$ and so it must be the case that $A\cap B\cap C=\emptyset$.
Since the above cases are exhaustive, $A\cap B\cap C=\emptyset$. Therefore if $C\subseteq A\Delta B$ then $A\cap B\cap C=\emptyset$.
From parts $(1)$ and $(2)$ we can conclude that if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.
$(\leftarrow)$ Suppose $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. Let $x$ be an arbitrary element of $C$. From $C\subseteq A\cup B$ and $x\in C$, $x\in A\cup B$. Now we consider two cases.
Case $1.$ Suppose $x\in A$. Now we consider two cases.
Case $1.1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\Delta B$.
Case $1.2.$ Suppose $x\notin A\setminus B$ and so $x\notin A$ or $x\in B$. Now we consider two cases.
Case $1.2.1.$ Suppose $x\notin A$ which is a contradiction.
Case $1.2.2.$ Suppose $x\in B$ which is a contradiction since $A\cap B\cap C=\emptyset$.
Since cases $1.2.1$ and $1.2.2$ lead to a contradiction then case $1.2$ leads to a contradiction. From case $1.1$ or case $1.2$ we can conclude $x\in A\Delta B$.
Case $2.$ Suppose $x\in B$ and a similar argument shows $x\in A\Delta B$.
Since case $1$ and case $2$ are exhaustive, $x\in A\Delta B$. Thus if $x\in C$ then $x\in A\Delta B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\Delta B)$ and so $C\subseteq A\Delta B$. Therefore if $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$ then $C\subseteq A\Delta B$.
From $(\rightarrow)$ and $(\leftarrow)$ we can conclude $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. $Q.E.D.$
Is my proof valid$?$ Is my proof unnecessarily redundant or every step is needed$?$
Thanks for your attention.
Best Answer
Your proof is correct. Here is a proof that avoids any mention of specific elements (following the theme of my answer to one of your previous questions). The key statements we use are the following:
(a) If $X$ and $Y$ are sets then $X \subseteq Y$ iff $X \setminus Y = \emptyset$.
(b) If $X$ and $Y$ are sets then $X \cup Y = \emptyset$ iff $X = \emptyset$ and $Y = \emptyset$.
(We discussed both of these before, so let's not reprove them!)
Now, in this problem we care about when $C \subseteq A \Delta B$. So, guided by property (a), we should examine $C \setminus (A\Delta B)$. Use axioms of set operations (e.g., De Morgan etc) to prove: $$ C \setminus (A\Delta B) = \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big)\tag{1} $$
I have hidden the proof of $(1)$ at the bottom of this answer; but try it yourself first. It's also a sensible thing to say out loud: $A \Delta B$ is the set of elements that are in either $A$ or $B$, but not both. So being in $C \setminus (A \Delta B)$ is the same as either being in $C$ and not in $A$ or $B$, or being in $C$ and in both $A$ and $B$.
Once you have $(1)$, the rest is very straightforward.
\begin{align} C \subseteq A \Delta B &\iff C \setminus (A \Delta B) = \emptyset \tag{using (a)} \\ &\iff \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big) = \emptyset \tag{using (1)}\\ &\iff C \setminus (A \cup B) = \emptyset \text{ and } A \cap B \cap C = \emptyset \tag{using (b)}\\ &\iff C \subseteq A\cup B \text{ and } A\cap B\cap C = \emptyset \tag{using (a)} \end{align}
Proof of $(1)$: