1123: Associativity for 3 Items Allows Any Association

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description/proof of that associativity for 3 items allows any association

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Topics

About: structure

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of structure.

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Target Context

• The reader will have a description and a proof of the proposition that for any structure, the associativity for any 3 items allows any association.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the structures }\}$
$\ast$: $\in \{\text{ the operations of }S\}$
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Statements:
$\mathrm{\forall }{s}_1,s_2,s_3\in S((s_1\ast s_2)\ast s_3=s_1\ast (s_2\ast s_3))$
$⟹$
$s_1\ast ...\ast s_n:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{n-1})\ast s_n$ can be associated in any way
//

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2: Note

Associativity is generally defined with respect to 3 items, which is being understood to allow any associativity. Let us confirm that that is indeed the case.

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3: Proof

Whole Strategy: Step 1: for each $1\le j\le n-1$, associate $s_j$ and $s_{j+1}$ 1st; Step 2: conclude the proposition.

Step 1:

$s_1\ast ...\ast s_n:=(...(((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Letting $a:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})$, it is $(...((a\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Applying the associativity for 3 items to $(a\ast s_j)\ast s_{j+1}$, $=(...(a\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n=(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n$.

So, any $s_j$ and $s_{j+1}$ can be associated.

Step 2:

Letting $b:=s_j\ast s_{j+1}$, it is $(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast b)\ast ...\ast s_{n-1})\ast s_n$.

By Step 1, its any neighboring 2 items can be associated.

That is what it means by $s_1\ast ...\ast s_n$ "can be associated in any way".

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References

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