Exercise 2.35#
It should satisfy the identity/unitality condition because \(S \cap X\) for any \(X\) that is a subset of \(S\) should be \(X\). See also Intersection (set theory); the intersection operation is associative, and commutative.
To see the monotonicity condition is satisfied we’ll use as an inference rule that a partially-applied \(\cap\) is order-preserving. That is, if we have that \(a_1 \leq a_2\) (that is, \(a_1 \subset a_2\)), it follows that \(a_1 \cap b \leq a_2 \cap b\). Then, starting from these two premises:
\[\begin{split}
\begin{align} \\
x_1 & \leq y_1 \\
x_2 & \leq y_2 \\
\end{align}
\end{split}\]
Applying \(\_ \cap y_2\) to the first premise:
\[\begin{split}
\begin{equation} \\
x_1 \cap y_2 \leq y_1 \cap y_2 \\
\label{eq:35a} \tag{a}
\end{equation}
\end{split}\]
Applying \(x_1 \cap \_\) to the second premise:
\[\begin{split}
\begin{equation} \\
x_1 \cap x_2 \leq x_1 \cap y_2 \\
\label{eq:35b} \tag{b}
\end{equation}
\end{split}\]
And combining \(\eqref{eq:35a}\) and \(\eqref{eq:35b}\) with transitivity:
\[
x_1 \cap x_2 \leq x_1 \cap y_2 \leq y_1 \cap y_2
\]