# General Associativity Theorem/Formulation 2

## Theorem

Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.

Let $\circ$ be an associative operation on $S$.

Let the set $P_n \left({a_1, a_2, \ldots, a_n}\right)$ be defined inductively by:

$P_1 \left({a_1}\right) = \left\{{a_1}\right\}$
$P_2 \left({a_1, a_2}\right) = \left\{{a_1 \circ a_2}\right\}$
$P_n \left({a_1, a_2, \ldots, a_n}\right) = \left\{{x \circ y: x \in P_r \left({a_1, a_2, \ldots, a_r}\right) \land y \in P_s \left({a_{r+1}, a_{r+2}, \ldots, a_{r+s}}\right), n = r+s}\right\}$

Then $P_n \left({a_1, a_2, \ldots, a_n}\right)$ consists of a unique entity which we can denote $a_1 \circ a_2 \circ \ldots \circ a_n$.

## Proof 1

The cases where $n = 1$ and $n = 2$ are clear.

Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.

If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction.

Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \circ y} = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ (again by induction).

If $r = 1$, then by induction $x \circ y = a_1 \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$.

Thus in either case, $x \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ which is a single element of $P_n$.

Hence we see that $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a single element.

$\blacksquare$

## Proof 2

Proof by strong induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$.

$P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$.

$P \left({2}\right)$ is the case:

$a_1 \circ a_2 = a_1 \circ a_2$

for which there is also nothing to prove.

### Basis for the Induction

$P \left({3}\right)$ is the case:

$\left({a_1 \circ a_2}\right) \circ a_3 = a_1 \circ \left({a_2 \circ a_3}\right)$

which is the definition of associativity.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k$.

Then we need to show:

The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k+1$.

### Induction Step

This is our induction step:

Consider the expressions:

$(1): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
$(2): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

for some $i, j \in \Z: 1 \le i, j \le k$.

We require to show that whatever the values of $i$ and $j$, these expressions are equal.

Without loss of generality, suppose that $i < j$.

Then the above expressions can be written:

$\left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

By the basis for the induction, the General Associativity Theorem holds for each of the parts in parenthesis.

Let:

$a = \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right)$
$b = \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right)$
$c = \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

By definition of associative operation:

$\left({a \circ b}\right) \circ c = a \circ \left({b \circ c}\right)$

This demonstrates the equality of the expressions $(1)$ and $(2)$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:

The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_n$ for $n \in \N$.

$\blacksquare$