This is Velleman's exercise 3.5.22.b:
Suppose $A$, $B$, and $C$ are sets. Prove that $A △ C ⊆ (A △ B) ∪ (B △ C)$.
Here's my proof of it:
First I prove "Theorem $\alpha$" which states that for sets $A$, $B$, and $C$, $A \setminus C ⊆ (A \setminus B) ∪ (B \setminus C)$.
Proof of "Theorem $\alpha$":
Proof. Suppose $x ∈ A\setminus C$, which means $x ∈ A$ but $x ∉ C$. Now we consider two different cases:
Case 1. $x ∈ B$. Then since $x ∉ C$, $x ∈ B\setminus C$ and ergo $x ∈ (A \setminus B) ∪ (B \setminus C)$.
Case 2. $x ∉ B$. Then since $x ∈ A$, $x ∈ A\setminus B$ and ergo $x ∈ (A \setminus B) ∪ (B \setminus C)$.
Since from both cases we have $x ∈ (A \setminus B) ∪ (B \setminus C)$, therefore $x ∈ A\setminus C \Rightarrow x ∈ (A \setminus B) ∪ (B \setminus C)$, which means $A △ C ⊆ (A △ B) ∪ (B △ C)$.
And now the proof of the original statement:
Proof. Suppose $x ∈ A △ C$, which means either $x ∈ A \setminus C$ or $x ∈ C \setminus A$ and hence we consider two different cases:
Case 1. $x ∈ A \setminus C$. Thus by "Theorem $\alpha$" we have either $x ∈ A \setminus B$ or $x ∈ B \setminus C$.
Subcase 1. $x ∈ A \setminus B$, from which we get $x ∈ A △ B$ and from which we'll have $x ∈ (A △ B) ∪ (B △ C)$.
Subcase 2. $x ∈ B \setminus C$, from which we get $x ∈ B △ C$ and from which we'll have $x ∈ (A △ B) ∪ (B △ C)$.
Case 2. $x ∈ C \setminus A$. Thus by "Theorem $\alpha$" we have either $x ∈ B \setminus A$ or $x ∈ C \setminus B$.
Subcase 1. $x ∈ B \setminus A$, from which we get $x ∈ A △ B$ and from which we'll have $x ∈ (A △ B) ∪ (B △ C)$.
Subcase 2. $x ∈ C \setminus B$, from which we get $x ∈ B △ C$ and from which we'll have $x ∈ (A △ B) ∪ (B △ C)$.
From both cases we have $x ∈ A △ C \Rightarrow x ∈ (A △ B) ∪ (B △ C)$ which is equivalent to $A △ C ⊆ (A △ B) ∪ (B △ C)$.
Is my proof correct? Could have I done it in any easier way?
Thanks in advance.
An easier way: $$A\triangle C=(A\triangle B)\triangle (B\triangle C) \subseteq (A\triangle B)\cup(B\triangle C)\,.$$