Does $x \in A$ and $x \notin B$ imply that $x \notin (A \cap B)$?

Question

For any two sets $A$ and $B$. Is it true that if $x \in A$ and $x \notin B$, then $x \notin (A \cap B)$?

Answer 1

It's true that $x \in A \land x \not\in B \implies x \not\in A \cap B$, but its converse is false: if $x \not\in A \cap B$, it's possible that $x \notin A \land x \notin B$, and this renders $x \in A \land x \not\in B$ false.

Answer 2

Two logical statements defining two sets in set theory are only equivalent if and only if the sets they define are equal. In the current question this is not so: $x\notin A\cap B$ even if $x\notin A$ and $x\in B$.

Answer 3

A bit more formal:

1) $x \in A$ and $x \not \in B$ $ \iff$

$x \in A$ and $x \in B^c$ $ \iff$

$x \in (A \cap B^c)$ $ \rightarrow$

$x \not \in A \cap B$.

2) $x \not \in (A \cap B)$ $ \iff$

$x \in A^c \cup B^c$ , i.e

$x \not \in A$ or $x \not \in B$.

Cf. Last line of Tampieri's answer.