Can't understand this set theory proof.

Question

I read the proof of a 'set theory' equation from a website called Meritnation. But I can't understand the proof after 30 minutes of trying and even find some mistakes in it.

This is the proof(I have copied it from a particular website and numbered the lines)

Now in line 5, it says $x \in (A \cap B)$ which is a subset to $(A \cup B) \cup (A-B).$ I think the $(A-B)$ is not required as $A \cup B$ already contains $(A-B).$ Same goes for line 10.

I can't understand the purpose of line 8 to 10 and line 11 comes out all of a sudden. From line 11, everything is easily understood and seems correct but almost all of the lines before it seem either wrong or unnecessary. Please help.

Answer 1

Lines 7-8: If $x$ is not in $A\cap B$, there are two overlapping subcases: $x\not\in A$ and/or $x\not\in B$.

Line 9: But the first subcase contradicts line 1, therefore certainly $x\not\in B$.

Line 10 should say that $x\in A,\ x\not\in B\implies x\in A-B$ which is $\subset (A\cap B)\cup(A-B)$.

Answer 2

This is not really an answer to your question, but here is an alternative proof, in a calculational style, which is a lot shorter and should be a lot simpler to follow.

So I'm posting this in the hope that it is helpful.

$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\followsfrom}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $Let's simplify $\;(A \cap B) \cup (A - B)\;$, by calculating which elements $\;x\;$ it contains: for all $\;x\;$,

$$\calc x \in (A \cap B) \cup (A - B) \op\equiv\hint{definition of $\;\cup\;$} x \in A \cap B \;\lor\; x \in A - B \op\equiv\hint{definition of $\;\cap\;$; definition of $\;-\;$} (x \in A \land x \in B) \;\lor\; (x \in A \land x \not\in B) \op\equiv\hint{logic: extract common conjunct, i.e., $\;\land\;$ distributes over $\;\lor\;$} x \in A \;\land\; (x \in B \lor x \not\in B) \op\equiv\hint{logic: excluded middle} x \in A \;\land\; \true \op\equiv\hint{logic: simplify} x \in A \endcalc$$

Therefore, by set extensionality, $\;(A \cap B) \cup (A - B) \;=\; A\;$.