Proof explanation $A\setminus B = A \setminus B \cap A$ $A\setminus B = A \setminus B \cap A$ Proof $A\setminus B \subseteq A \setminus B \cap A$: $x\in A\setminus B \to x\in A \land x \notin B$ $x\in A \land (x \notin B \color{red}{\lor x\notin A})$ $x \in A \land (x\notin B \cap A)$ $x \in A\setminus B\cap A$ Proof $A \setminus B \cap A \subseteq A\setminus B$: $x \in A \setminus B \cap A \to x\in A \land x\notin B \cap A$ $x \in A \land (x\notin B \color{red}{\lor x\notin A})$ $x \in A \land x \notin B$ $x \in A \setminus B $ Therefore $A\setminus B = A \setminus B \cap A$ I am confused how the stuff in the red is formed. Can we add an or statement to anything and ignore if its true or not if the first part of the or is true?

I'd write the proof in this way

$x\in A\setminus B \Rightarrow x\in A \land x\
otin B\Rightarrow x\in A \land (x\
otin A\lor x\
otin B)\\\\\Rightarrow x\in A\land \
eg(x\in A\land x\in B)$ by De Morgan's law

$ x\in A\land x\
otin A\cap B\Rightarrow x\in A\setminus(A\cap B) \Rightarrow A\setminus B \subseteq A \setminus (B \cap A)$

$x\in A \setminus (B \cap A)\Rightarrow x\in A \land x\
otin (A\cap B) \Rightarrow x\in A \land \
eg (x\in A \land x\in B)\\\ \Rightarrow x\in A \land x\
otin A \lor x\
otin B$ again for De Morgan's

$x\in A \land x\
otin B\Rightarrow x\in A\setminus B\Rightarrow A \setminus (B \cap A)\subseteq A\setminus B$

Therefore

$A\setminus B = A \setminus (B \cap A)$