Preimage of Serial Relation is Domain
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Theorem
Let $\RR$ be a serial relation on $S$.
Then the preimage of $\RR$ is $S$ (the domain of $\RR$).
Proof
| \(\ds S\) | \(\supseteq\) | \(\ds \Preimg \RR\) | Definition of Preimage of Relation | |||||||||||
| \(\ds \forall s \in S: \exists t \in S: \, \) | \(\ds \tuple {s, t}\) | \(\in\) | \(\ds \RR\) | Definition of Serial Relation | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall s \in S: \, \) | \(\ds s\) | \(\in\) | \(\ds \Preimg \RR\) | Definition of Preimage of Relation | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds \Preimg \RR\) | Definition of Subset | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(=\) | \(\ds \Preimg \RR\) | Definition 2 of Set Equality |
$\blacksquare$