Intersection with Subclass is Subclass

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Theorem

Let $A$ and $B$ be classes.


Then:

$A \subseteq B \iff A \cap B = A$

where:

$A \subseteq B$ denotes that $A$ is a subclass of $B$
$A \cap B$ denotes the intersection of $A$ and $B$.


Proof

Let $A \cap B = A$.

Then by the definition of class equality:

$A \subseteq A \cap B$

Thus:

\(\ds x\) \(\in\) \(\ds A\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A \cap B\) Definition of Subclass: $A \subseteq A \cap B$
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A\) Definition of Class Intersection
\(\, \ds \land \, \) \(\ds x\) \(\in\) \(\ds B\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds B\) Definition of Conjunction
\(\ds \leadsto \ \ \) \(\ds A\) \(\subseteq\) \(\ds B\) Definition of Subclass

$\Box$


Now let $A \subseteq B$.

\(\ds x\) \(\in\) \(\ds A \cap B\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A\) Definition of Class Intersection
\(\, \ds \land \, \) \(\ds x\) \(\in\) \(\ds B\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A\) Definition of Conjunction
\(\ds \leadsto \ \ \) \(\ds A \cap B\) \(\subseteq\) \(\ds A\) Definition of Subclass


We also have:


\(\ds x\) \(\in\) \(\ds A\)
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A\)
\(\, \ds \land \, \) \(\ds x\) \(\in\) \(\ds B\) Definition of Subclass
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds A \cap B\) Definition of Class Intersection



So as we have:

\(\ds A \cap B\) \(\subseteq\) \(\ds A\)
\(\ds A\) \(\subseteq\) \(\ds A \cap B\)

it follows from the definition of class equality that:

$A \cap B = A$

$\blacksquare$


Also see


Sources

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