Prove $(A \cup B) \setminus (C \setminus B) = B \cup(A \setminus C)$

elementary-set-theorysolution-verification

Let $A,B,C$ be sets.

I am required to prove the following statement:

$$(A \cup B) \setminus (C \setminus B) = B \cup(A \setminus C)$$

First, I tried to write a set-wise proof, but didn't manage to do so. I would like to see such proof.

Second, I tried to prove it element-wise. Here's my try:

From the left side, I can write $x \in A\cup B$ and $x \notin C \setminus B$. This splits into $2$ cases:

Case 1: $x \in A \iff x \in A\setminus C \iff x \in B \cup(A \setminus C)$

Case 2: $x \in B \iff x \in B \cup(A \setminus C)$

Since both cases use only "iff" statements, and both imply that element from one side belongs to the other side as well, then both sides are equal, because it proves both directions needed. But this "proof" completely ignores $x \notin C \setminus B$ so I doubt its validity.

  1. So, is my proof correct? Where am I wrong and how do I make use of $x \notin C \setminus B$ (if it's even needed).
  2. Can this be proven group-wise?

Best Answer

Proof:

We have:

$x \in (A \cup B) \setminus (C \setminus B)$

iff $x \in (A \cup B) \land x \notin (C \setminus B)$

Then, $x \notin (C \setminus B)$, iff $\neg (x \in (C \setminus B))$, iff $\neg (x \in C \land x \notin B)$, iff $x \notin C \lor x \in B$.

So: $x \in (A \cup B) \land (x \notin (C \setminus B))$

iff $x \in (A \cup B) \land (x \notin C \lor x \in B)$

iff $[x \in (A \cup B) \land x \notin C] \lor [x \in (A \cup B) \land x \in B]$.

Since we always have $B \subseteq (A \cup B)$, we thus have [$x \in (A \cup B) \land x \in B$] iff $x \in B$.

Also, $[x \in (A \cup B) \land x \notin C]$, iff $[(x \in A \land x \notin C) \lor (x \in B \land x \notin C)]$, iff $[(x \in A \setminus C) \lor (x \in B \setminus C)]$.

Therefore: $[x \in (A \cup B) \land x \notin C] \lor [x \in (A \cup B) \land x \in B]$

iff $(x \in A \setminus C) \lor (x \in B \setminus C) \lor x \in B$

Since we always have $(B \setminus C) \subseteq B$, then we have:

$(x \in A \setminus C) \lor (x \in B \setminus C) \lor x \in B$

iff $(x \in A \setminus C) \lor x \in B$

iff $x \in (A \setminus C) \cup B$

Therefore, $x \in (A \cup B) \setminus (C \setminus B)$ iff $x \in (A \setminus C) \cup B$, and this means $(A \cup B) \setminus (C \setminus B) = (A \setminus C) \cup B$.

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