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
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(d)}$