Associativity on Four Elements

Theorem
Let $$\left({S, \circ}\right)$$ 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 $$\left({S, \circ}\right)$$ is a semigroup:
 * it is closed;
 * $$\circ$$ is associative.

It can be shown that 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:
 * $$\forall s_1, s_2, s_3 \in S: \left({s_1 \circ s_2}\right) \circ s_3 = s_1 \circ \left({s_2 \circ s_3}\right)$$

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

So:

$$ $$ $$ $$

Comment
This entry is clearly superseded by the General Associativity Theorem, which demonstrates this rule for any number of elements.

The entry is included, however, for its instructive and illustrative nature.