Let's say we have three set's $A, B, C$ with elements $\in \Bbb Z$ . If there is an operation
$A \setminus (A\cap (B \setminus C)^\mathsf{c})\cup(B\cap C)$
let $x=(A\cap (B\setminus C)^\mathsf{c})$ and $y=(B\cap C)$ , from the above operation
$\therefore A \setminus x\cup y$
Now my question is what operation must we go for first $A\setminus x$ or $x\cup y$, as they might result in different answers?
As there is a BODMAS rule in Algebra, is there similar kind of rule in set's operations also?
PS: $X\setminus Y \Rightarrow X-Y$
The Khan Academy video where he does a question using this relation. Just for the sake of example.
Just as arithmetic features binary operations on numbers, set theory features binary operations on sets. The binary operations are Idempotent, Commutative, Associative, Distributive, etc.
In elementary set theory, the three binary operations: union, intersection, and difference are all equal in the order of precedence. The three binary operation (known as basic operations) have a higher precedence than other set operations such as symmetric difference or cartesian product.
From the example given in the question (or the Khan Academy video), $(A \setminus x)\cup y$ = $A \setminus (x\cup y)$. Let us verify this with the example; $$A = \{3, 7, -5, 0, 13\}$$ $$B = \{0, 17, 3, 9, 19\}$$ $$C = \{18, 19, 3, 17\}$$
Let us consider $x=(A\cap (B\setminus C)^\mathsf{c}) =\{3, 7, -5, 13\}$ and $y=(B\cap C)=\{17, 3, 19\}$. Clearly, $(A \setminus x)\cup y$ = $A \setminus (x\cup y)=\{0, 17, 3, 19\}$.
See Algebra of sets: https://en.wikipedia.org/wiki/Algebra_of_sets