Extended Transitivity

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be a set.

Let $\mathcal R$ be a transitive relation on $S$.

Let $\mathcal R^=$ be the reflexive closure of $\mathcal R$.

Let $a, b, c \in S$.


Then:

\((1):\quad\) \(\displaystyle \left({ a \mathrel{\mathcal R} b }\right) \land \left({ b \mathrel{\mathcal R} c }\right)\) \(\implies\) \(\displaystyle a \mathrel{\mathcal R} c\)
\((2):\quad\) \(\displaystyle \left({ a \mathrel{\mathcal R} b }\right) \land \left({ b \mathrel{\mathcal R^=} c }\right)\) \(\implies\) \(\displaystyle a \mathrel{\mathcal R} c\)
\((3):\quad\) \(\displaystyle \left({ a \mathrel{\mathcal R^=} b }\right) \land \left({ b \mathrel{\mathcal R} c }\right)\) \(\implies\) \(\displaystyle a \mathrel{\mathcal R} c\)
\((4):\quad\) \(\displaystyle \left({ a \mathrel{\mathcal R^=} b }\right) \land \left({ b \mathrel{\mathcal R^=} c }\right)\) \(\implies\) \(\displaystyle a \mathrel{\mathcal R^=} c\)


Proof

$(1)$ follows from the definition of a transitive relation.

$(4)$ follows from Reflexive Closure of Transitive Relation is Transitive.

Suppose that:

$\left({ a \mathrel{\mathcal R} b }\right) \land \left({ a \mathrel{\mathcal R^=} c }\right)$

By the definition of reflexive closure:

$b \mathrel{\mathcal R} c$ or $b = c$

If $b=c$, then since a $\mathrel{\mathcal R} b$:

$a \mathrel{\mathcal R} c$

If $a \mathrel{\mathcal R} c$ then by transitivity of $\mathcal R$:

$a \mathrel{\mathcal R} c$

Thus $(2)$ holds.

A similar arguments proves that $(3)$ holds as well.

$\blacksquare$