Construction of Transitive Closure of Relation

Theorem

Let $\RR$ be a relation.

Let $\RR^+$ be the relation which is constructed from $\RR$ as follows:

$(1): \quad$ If $\tuple {a, b} \in \RR$, then $\tuple {a, b} \in \RR^+$

$(2): \quad$ If $\tuple {a, b} \in \RR^+$ and $\tuple {b, c} \in \RR$, then $\tuple {a, c} \in \RR^+$

$(3): \quad$ Nothing is in $\RR^+$ unless it so follows from $(1)$ and $(2)$.

Then $\RR^+$ is the transitive closure of $\RR$.

Proof

Let $\tuple {x, y} \in \R^+$ from rules $(1)$ and $(2)$.

Then either:

$\tuple {x, y}$ belongs there because $\RR \subseteq \RR^+$

or:

$\tuple {x, y}$ belongs there, because if it were not then $\RR^+$ would not be transitive.

It remains to be shown that $\RR^+$ is in fact transitive.

This needs considerable tedious hard slog to complete it. In particular: "Easy inductive proof", says H&U, that $\RR^+$ is in fact transitive To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Because of the method of construction, it follows that $\RR^+$ is the smallest transitive relation on $S$ which contains $\RR$ as a subset.

Sources

• 1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: $1.5$ Relations: Closures of Relations