Set Difference of Doubleton and Singleton is Singleton
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Theorem
Let $x, y$ be distinct objects.
Then:
- $\set{x, y} \setminus \set x = \set y$
Proof
| \(\ds \set {x, y} \setminus \set x\) | \(=\) | \(\ds \set {z: z \in \set {x, y} \land z \notin \set x}\) | Definition of Set Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z: \paren {z = x \lor z = y} \land z \notin \set x}\) | Definition of Doubleton | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z: \paren {z = x \lor z = y} \land z \ne x}\) | Definition of Singleton | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z: \paren {z = x \land z \ne x} \lor \paren {z = y \land z \ne x} }\) | Conjunction Distributes over Disjunction | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z: \bot \lor \paren {z = y \land z \ne x} }\) | Definition of Contradiction | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z: \paren {z = y \land z \ne x} }\) | Disjunction with Contradiction | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set {z : z = y}\) | Rule of Simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set y\) | Definition of Singleton |
$\blacksquare$