# Associativity on Four Elements

## Theorem

Let $\struct {S, \circ}$ be a semigroup.

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

Then:

$a \circ b \circ c \circ d$

gives a unique answer no matter how the elements are associated.

## Proof

As $\struct {S, \circ}$ is a semigroup:

it is closed
$\circ$ is associative

From Parenthesization of Word of $4$ Elements, there are exactly $5$ different ways of inserting brackets in the expression $a \circ b \circ c \circ d$.

As $\circ$ is associative, we have that:

$\forall s_1, s_2, s_3 \in S: \paren {s_1 \circ s_2} \circ s_3 = s_1 \circ \paren {s_2 \circ s_3}$

As $\struct {S, \circ}$ is closed, we know that all products of elements from $\set {a, b, c, d}$ are in $S$, and are likewise bound by the associativity of $S$.

So:

 $\ds \paren {\paren {a \circ b} \circ c} \circ d$ $=$ $\ds \paren {a \circ \paren {b \circ c} } \circ d$ $\ds$ $=$ $\ds a \circ \paren {\paren {b \circ c} \circ d}$ $\ds$ $=$ $\ds a \circ \paren {b \circ \paren {c \circ d} }$ $\ds$ $=$ $\ds \paren {a \circ b} \circ \paren {c \circ d}$

$\blacksquare$

## Sources

in the context of addition