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Possible Duplicate:
How does one actually show from associativity that one can drop parentheses?

Claim: Let $(S, *)$ be a semigroup and $a_1, \ldots, a_n \in S$. Then $a_1 * \ldots * a_n$ is unique.

I'm having trouble proving this claim, because I'm a beginner, I'm sure, and would like some guidance if you please. Here's my attempt so far:

Edit: Following AndréNicolas advice, here's my second attempt.

Proof: My strategy is the following. Given a set S of elements:

  1. Define the set of all products;
  2. Define the set of all normalized products;
  3. Show how we can normalize all products;
  4. Show that all products are equal to their normalized version.

Step 1: Define the set of all products:

Let $*_S$, the set of all products over $S$, be defined inductively as follows:

  • $a \in *_S$, if $a \in S$;
  • $(a * x) \in *_S$, if $a \in S$ and $x \in *_S$;
  • $(x * a) \in *_S$, if $a \in S$ and $x \in *_S$

Step 2: Define the set of all normalized products:

Let $\star_S$, the set of all normalized products over $S$, be defined inductively as follows:

  • $a \in \star_S$, if $a \in S$;
  • $(a * x) \in \star_S$, if $a \in S$ and $x \in \star_S$

Step 3: Show how we can normalize all products:

Let $\searrow_S: *_S \to \star_S$ be the following application, also defined inductively:

$\begin{array}{rcl} a & \mapsto & a \\ (a * x) & \mapsto & (a * \searrow_S(x)) \\ (x * a) & \mapsto & \begin{cases} (a\prime * a) &, \searrow_S(x) = a\prime \\ (a\prime * \searrow_S(y * a)) & , \searrow_S(x) = (a\prime * y) \end{cases} \end{array}$

Step 4: Show that all products are equal to their normalized version:

This proceeds by strong induction over $*_S$. Let $\Phi(x) \equiv x = \searrow_S(x)$.

  • $a = \searrow_S(a) \equiv a = a$;
  • Suppose $x = \searrow_S(x)$. It's left to show that $(a * x) = \searrow_S(a * x) \equiv (a * x) = (a * \searrow_S(x)) \equiv (a * x) = (a * x)$.
  • Suppose $x = \searrow_S(x)$. It's left to show that $(x * a) = \searrow_S(x * a)$. There are two cases:
    • $\searrow_S(x) = a\prime$. Then $(x * a) = \searrow_S(x * a) \equiv (a\prime * a) = (a\prime * a)$.
    • $\searrow_S(x) = (a\prime * y)$. Then $(x * a) = \searrow_S(x * a) \equiv ((a\prime * y) * a) = (a\prime * \searrow_S(y * a))$. and this where I am stuck now. I am not sure how strong induction plays, in order to develop this further. Can you please criticise what I've done so far?

Thanks for taking the time to read this question of mine. Best wishes,

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William T.
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  • Working with no parentheses may be confusing. Define "normal form" by saying it is something like $a(b(c(de)))$ (all left parentheses pushed as far left as possible). Show by (strong) induction that any product is equal to a normal form product. – André Nicolas Nov 19 '11 at 17:08
  • @AndréNicolas: Thanks, I will try that and come back later to report on my findings. – William T. Nov 19 '11 at 17:56
  • The claim here comes as a little more clearly stated in my opinion than in the linked thread of Arturo's, but I do agree that it seems like a duplicate. – Doug Spoonwood Nov 20 '11 at 01:11

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