# Extended Transitivity

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## 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:

\((1):\quad\) | \(\displaystyle \paren{ a \mathrel {\RR} b } \land \paren { b \mathrel {\RR} c }\) | \(\implies\) | \(\displaystyle a \mathrel{\RR} c\) | ||||||||||

\((2):\quad\) | \(\displaystyle \paren { a \mathrel{\RR} b } \land \paren { b \mathrel{\RR^=} c }\) | \(\implies\) | \(\displaystyle a \mathrel{\RR} c\) | ||||||||||

\((3):\quad\) | \(\displaystyle \paren { a \mathrel{\RR^=} b } \land \paren { b \mathrel{\RR} c }\) | \(\implies\) | \(\displaystyle a \mathrel{\RR} c\) | ||||||||||

\((4):\quad\) | \(\displaystyle \paren { a \mathrel{\RR^=} b } \land \paren { b \mathrel{\RR^=} c }\) | \(\implies\) | \(\displaystyle a \mathrel{\RR^=} c\) |

## Proof

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

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

Suppose that:

- $\paren { a \mathrel{\RR} b } \land \paren { b \mathrel{\RR^=} c }$

By the definition of reflexive closure:

- $b \mathrel{\RR} c$ or $b = c$

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

- $a \mathrel{\RR} c$

If $b \mathrel{\RR} c$ then by transitivity of $\RR$:

- $a \mathrel{\RR} c$

Thus $(2)$ holds.

A similar argument proves that $(3)$ holds as well:

Suppose that:

- $\paren { a \mathrel{\RR^=} b } \land \paren { b \mathrel{\RR} c }$

By the definition of reflexive closure:

- $a \mathrel{\RR} b$ or $a = b$

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

- $a \mathrel{\RR} c$

If $a \mathrel{\RR} b$ then by transitivity of $\RR$:

- $a \mathrel{\RR} c$

Thus $(3)$ holds.

$\blacksquare$