Class Difference of B with Class Difference of A with B

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Theorem

Let $A$ and $B$ be classes.


Then:

$B \setminus \paren {A \setminus B} = B$

where $S \setminus T$ denotes class difference.


Proof

\(\ds \) \(\) \(\ds x \in B \setminus \paren {A \setminus B}\)
\(\ds \) \(\leadstoandfrom\) \(\ds x \in B \land x \notin \paren {A \setminus B}\) Definition of Class Difference
\(\ds \) \(\leadstoandfrom\) \(\ds x \in B \land \paren {x \notin A \lor x \in B}\) De Morgan's Laws
\(\ds \) \(\leadstoandfrom\) \(\ds \paren {x \in B \land x \notin A} \lor \paren {x \in B \land x \in B}\) Conjunction is Left Distributive over Disjunction
\(\ds \) \(\leadstoandfrom\) \(\ds \paren {x \in B \land x \notin A} \lor x \in B\) Rule of Idempotence: Conjunction
\(\ds \) \(\leadstoandfrom\) \(\ds x \in B\) Disjunction Absorbs Conjunction
\(\ds \) \(\leadsto\) \(\ds B \setminus \paren {A \setminus B} = B\) Definition of Class Equality

$\blacksquare$


Sources