Transitive Closure (Relation Theory)/Examples/Arbitrary Example 2
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Example of Transitive Closure
Let $S = \set {1, 2, 3, 4, 5}$ be a set.
Let $\RR$ be the relation on $S$ defined as:
- $\RR = \set {\tuple {1, 2}, \tuple {2, 3}, \tuple {3, 4}, \tuple {5, 4} }$
The transitive closure $\RR^+$ of $\RR$ is given by:
- $\RR^+ = \set {\tuple {1, 2}, \tuple {1, 3}, \tuple {1, 4}, \tuple {2, 3}, \tuple {2, 4}, \tuple {3, 4}, \tuple {5, 4} }$
Proof
By Construction of Transitive Closure of Relation:
\(\ds \tuple {1, 2}\) | \(\in\) | \(\ds \RR^+\) | Rule $(1)$ | |||||||||||
\(\ds \tuple {2, 3}\) | \(\in\) | \(\ds \RR^+\) | Rule $(1)$ | |||||||||||
\(\ds \tuple {3, 4}\) | \(\in\) | \(\ds \RR^+\) | Rule $(1)$ | |||||||||||
\(\ds \tuple {5, 4}\) | \(\in\) | \(\ds \RR^+\) | Rule $(1)$ |
That completes the phase of the construction which uses Rule $(1)$.
Then:
\(\ds \tuple {1, 2}\) | \(\in\) | \(\ds \RR\) | by hypothesis | |||||||||||
\(\ds \tuple {2, 3}\) | \(\in\) | \(\ds \RR^+\) | a priori | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {1, 3}\) | \(\in\) | \(\ds \RR^+\) | Rule $(2)$ |
\(\ds \tuple {2, 3}\) | \(\in\) | \(\ds \RR\) | by hypothesis | |||||||||||
\(\ds \tuple {3, 4}\) | \(\in\) | \(\ds \RR^+\) | a priori | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {2, 4}\) | \(\in\) | \(\ds \RR^+\) | Rule $(2)$ |
\(\ds \tuple {1, 2}\) | \(\in\) | \(\ds \RR\) | by hypothesis | |||||||||||
\(\ds \tuple {2, 4}\) | \(\in\) | \(\ds \RR^+\) | a priori | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {1, 4}\) | \(\in\) | \(\ds \RR^+\) | Rule $(2)$ |
There are no further elements of $\RR$ for which there are corresponding elements of $\RR^*$ that satisfy Rule $(2)$.
Hence the result.
$\blacksquare$
Sources
- 1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: Exercises: $1.7$