Deriving intersection of two truth sets

Question

$ \newcommand{\Set}[2]{% \{\, #1 \mid #2 \, \}% }$

Let $A = \Set{x}{P(x)}$, $B = \Set{x}{Q(x)}$. We know by definition that

$$ A \setminus B = \Set{x}{P(x) \land \lnot Q(x)}.$$

Since $X \cap Y = X \setminus (X \setminus Y)$ for two sets $X$ and $Y$ then following should hold true.

\begin{align} & A \cap B = \\ & A \setminus (A \setminus B) = \\ & \Set{x}{P(x)} \setminus (\Set{x}{P(x) \land \lnot Q(x)}) = \\ &\Set{x}{P(x) \land \lnot(P(x) \land \lnot Q(x))} = \\ & \Set{x}{P(x) \land \lnot P(x) \lor Q(x)} = \Set{x}{Q(x)} \end{align}

Where is the mistake? The answer should be $\Set{x}{P(x) \land Q(x)}$. Is the intersection formula wrong or is it not generally true that

$$ \Set{x}{\varphi(x)} \setminus \Set{x}{\psi(x)} = \Set{x}{\varphi(x) \land \lnot \psi(x)} $$

for sentences $\varphi(x)$ and $\psi(x)$?

Answer 1

The error originated in the line preceding "Where is the mistake?"; you had crucially left out a pair of parentheses. So, correction: $ \newcommand{\Set}[2]{% \{\, #1 \mid #2 \, \}% }$ \begin{align} & A \setminus (A \setminus B) \\ = &\Set{x}{P(x) \land \lnot(P(x) \land \lnot Q(x))} \\ = & \Set{x}{P(x) \land (\lnot P(x) \lor Q(x))}\\ = & \Set{x}{(P(x) \land \lnot P(x)) \lor (P(x) \land Q(x))}\\ = & \Set{x}{P(x) \land Q(x)}. \end{align}