# is this proof wrong pertaining to Ordered Pairs?

I was understanding this proof :

(A-B) X C = (A X B) - (B X C)

A transition in statements in proof seemed so wrong to me , which are following :

$(x \in A \;and\; x \notin B) \;and \; y \in C$

$(x \in A\; and \;y \in C) \;and\; (x \notin B\; and \;y \in C)$

$(x,y) \in (A X C) \;and\; (x,y) \notin (B X C)$

my question lies in last two statements , how can we write $(x\notin B\;and\;y \in C)$ to $(x,y) \notin (B X C)$ ?

I mean statement $(x \notin B \;and\; y \in C)$ says x doesn't belongs to B and y belongs to C.

It's equivalent (in proof) statement says (x,y) such that x doesn't belongs to B and y doesn't belongs to C , isn't that a contradiction or am I missing something here ?

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$B\times C$ is the set of all ordered pairs $\langle b,c\rangle$ such that $b\in B$ and $c\in C$. If $x\notin B$, the pair $\langle x,y\rangle$ cannot be in $B\times C$, because its first component, $x$, is not in $B$. it doesn’t matter whether $y\in C$ or $y\notin C$: the fact that $x\notin B$ is already enough to guarantee that $\langle x,y\rangle\notin B\times C$.
As an example, suppose that $B=\{1,2\}$ and $C=\{3,4,5\}$. Then $\langle 3,4\rangle\notin B\times C$, because $3\notin B$; $\langle 1,6\rangle\notin B\times C$, because $6\notin C$; $\langle 1,4\rangle\in B\times C$ because $1\in B$ and $4\in C$; and $\langle 3,6\rangle\notin B\times C$ both because $3\notin B$ and because $6\notin C$: it fails for two reasons to belong to $B\times C$.