Section I.III: Words

Word Axioms

Important

To reiterate the introduction to this section, the current formal system does not seek to describe a generative grammar. Its theorems cannot be used as schema for generating grammatical sentences. The intent of this analysis is to treat Words as interpretted constructs embedded in a syntactical structure that is independent of their specific interpretations.

A Word is a type of String constructed through concatenation that has been assigned by semantic content. A Language is the aggregate of all Words.

\[\forall \alpha \in L: \alpha \in S\]

Or equivalently,

\[L \subset S\]

Axiom V: Measure

No Words have a String Length of 0.

\[\forall \alpha \in L: l(\alpha) \neq 0\]

Axiom VI: Discovery

No Character in a Word can be a Delimiter.

\[\forall \alpha \in L: \forall i \in N_{l(\alpha)}: \alpha[i] \neq \sigma\]

Axiom VII: Canonization

All Words are canonical.

\[\forall \alpha \in L: \alpha \in \mathbb{S}\]

Classes

Definition 1.3.1: Reflective Words

Let \(\alpha \in L\). \(\alpha\) belongs to the set of Reflective Words, denoted \(R\), if it satisfies the open formula,

\[(\alpha \in R) \equiv (\alpha = {\alpha}^{-1})\]

A Word will be referred to as reflective if it belongs to the class of Reflective Words.

Note

\(R\) may be defined equivalently through set builder notation,

\[R = \{ \alpha \in L \mid \alpha = {\alpha}^{-1} \}\]

Example The following table lists some reflective English words.

Word
mom
dad
noon
racecar
madam
level
civic

∎

Definition 1.3.2: Invertible Words

Let \(\alpha \in L\). \(\alpha\) belongs to the set of Invertible Words, denoted \(I\), if it satisfies the open formula,

\[(\alpha \in I) \equiv ({\alpha}^{-1} \in L)\]

A Word will be referred to as invertible if it belongs to the class of Invertible Words.

Important

A Word is invertible if and only if its inverse belongs to the Language.

Example The following table lists some English words and their inverses (where applicable).

Word	Inverse
time	emit
saw	was
raw	war
dog	god
pool	loop
cat	x
you	x
help	x
door	x
book	x

∎

Note

Invertible Words are often called semiordnilaps in other fields of study.

Theorem 1.3.1

A Word is invertible if and only if its inverse is invertible.

\[\forall \alpha \in L: \alpha \in I \equiv {\alpha}^{-1} \in I\]

Proof Let \(\alpha \in L\).

(\(\rightarrow\)) Assume \(\alpha \in I\). By the definition of invertible Words,

\[{\alpha}^{-1} \in L\]

By Theorem 1.2.9,

\[({\alpha}^{-1})^{-1} = \alpha\]

Therefore, by assumption,

\[({\alpha}^{-1})^{-1} \in L\]

By the definition of invertible Words,

\[{\alpha}^{-1} \in I\]

(\(\leftarrow\)) Assume \({\alpha}^{-1} \in L\) such that \({\alpha}^{-1} \in I\). By the definition of invertible Words,

\[({\alpha}^{-1})^{-1} \in L\]

By Theorem 1.2.9,

\[\alpha \in L\]

Since \({\alpha}^{-1} \in L\) by assumption, it follows immediately from the definition of invertible Words,

\[\alpha \in I\]

Summarizing and generalizing,

\[\forall \alpha \in L: \alpha \in I \equiv {\alpha}^{-1} \in I\]

∎

Theorem 1.3.2

Reflective Words are a subset of Invertible Words.

\[R \subset I\]

Proof Let \(\alpha in R\) and \(l(\alpha) = n\). By the definition of Reflective Words,

\[\alpha = \alpha^{-1}\]

Since \(\alpha \in L\) by assumption, it follows \(\alpha in I\). In other words,

\[\alpha \in R \implies \alpha \in I\]

But this is exactly the definition of the subset relation in set theory, therefore,

\[R \subset I\]

∎

Limitations

Definition 1.3.3: Phrases

Let \(n \in \mathbb{N}\). A Phrase of Word Length \(n\), denoted \(P_n\), is defined as an ordered sequence of \(n\) Words, not necessarily distinct,

\[P_n = \{ (i, \alpha_i) \mid (i \in N_n) \land (\alpha \in L) \}\]

\[P_n = (\alpha_1, \alpha_2, ..., \alpha_n)\]

where each \(\alpha_i \in L\). If \(1 \leq i \leq n\), \(P_n(i)\) denotes the Word \(\alpha_a\) at index \(i\) of the Phrase, so the Phrase may be written,

\[P_n = (P_n(1), P_n(2), ... P_n(n))\]

When \(n = 0\), a Phrase is defined to be \(\varnothing\),

\[P_0 = \varnothing\]

Definition 1.3.4: Lexicons

Let \(n \in \mathbb{N}\). A Language’s \(n^{\text{th}}\) Lexicon, denoted \(L_n\), is defined as the set of all Phrases of length \(n\),

\[L_n = \{ p \mid \forall p: p = P_n \}\]

Defintion 1.3.5: Limitation

Let \(p \in L_n\). The Limitation of \(p\), denoted \(\Pi_{i=1}^{n} p(i)\) is defined inductively using the following schema,

• Empty: \(\Pi_{i=1}^{0} p(i) = \varepsilon\)

• Basis: \(\Pi_{i=1}^{1} p(i) = \alpha_1\)

• Induction: \(\Pi_{i=1}^{n} p(i) = (\Pi_{i=1}^{n-1} p(i))(\sigma)(\alpha_n)\)

The process of Limitation, \(\Pi_{i=1}^{n} p(i)\), will be referred to as “delimiting” a Phrase or Words.

Note

A Limitation, though notationally complex, can be understood as shorthand for the iterated concatenation of words and Delimiters. is the presence of the Delimiter in the Induction clause. In other words, a Limitation inserts Delimiters inbetween each Word in the Lexicon over which the index is ranging.

Example Let \(L = L_{\text{english}}\). Consider calculating the Limitation of the following Phrase,

\[P_3 = (\text{mother}, \text{may}, \text{i})\]

Apply the Basis clause Limitations ,

\[n = 1: \quad \Pi_{i=1}^{1} \alpha_i = \text{mother}\]

The Limitation can then be built up recursively using the Induction clause,

\[n = 2: \quad \Pi_{i=1}^{2} \alpha_i = (\Pi_{i=1}^{1} \alpha_i)(\sigma)(\text{may})= (\text{mother})(\sigma\text{may}) = \text{mother}\sigma\text{may}\]

\[n = 3: \quad \Pi_{i=1}^{3} \alpha_i = (\Pi_{i=1}^{2} \alpha_i)(\sigma)(\text{i}) = (\text{mother}\sigma\text{may})(\sigma\text{i}) = \text{mother}\sigma\text{may}\sigma\text{i}\]

So the Limitation of the Phrase is shown to be,

\[\Pi_{i=1}^{3} \alpha_i = \text{mother may I}\]

Important

The result of a Limitation is a String. Since a Limitation is shorthand for alternating concatenation of Characters and Delimiters, the closure of Limitations over \(S\) is guaranteed by the closure of concatenation over \(S\)

∎

Theorem 1.3.3

All Limitations are unique within the Canon.

\[\forall n \in \mathbb{N}: \forall p \in L_n: \exists! s \in \mathbb{S}: s = \Pi_{i=1}^{n} p(i)\]

Proof Let \(n \in \mathbb{N}\) and \(p \in L_n\) such that,

\[p = (\alpha_1, \alpha_2, ..., \alpha_n)\]

The proof will proceed by induction on \(n\).

Basis: Assume \(n = 1\). By Basis clause of Limitations,

\[\Pi_{i=1}^{1} p(i) = \alpha_1\]

Induction: Assume for \(k \geq 1\), these exists a unique String \(s_k\) such that,

\[s_k = \Pi_{i=1}^{k} p(i)\]

By Induction clause of Limitations,

\[\Pi_{i=1}^{k+1} p(i) = (\Pi_{i=1}^{k} p(i))(\sigma)(\alpha_{k+1})\]

By inductive hypothesis,

\[s_{k+1} = \Pi_{i=1}^{k+1} p(i) = ({s_k})(\sigma)(\alpha_{k+1})\]

Therefore, by induction,

\[\forall n \in \mathbb{N}: \forall p \in L_n: \exists! s \in \mathbb{S}: s = \Pi_{i=1}^{n} p(i)\]

∎

This subsection closes with a definition that will be used to quantify a theorem regarding Word Length.

Definition 1.3.6: Dialect

Let \(L_i\) be the \(i^{\text{th}}\) Lexicon of Language \(L\). The Language’s Dialect, denoted \(D\), is defined as the set,

\[D = \bigcup_{i=1}^{\infty} \{ s \in S \mid \exists p \in L_i: s = \Pi_{j=1}^{i} p(j) \}\]

Warning

The type of each set defined in this section should be carefully analyzed.

• A Phrase is an ordered set of Words.

• A Lexicon is the set of all Phrases of a fixed Word Length.

• A Dialect is the set of Strings formed by delimiting every Phrase in every Lexicon of a Language.

Example Let \(L = \{ \text{hakuna}, \text{matata} \}\). Then, the first few Lexicons are given below,

\[\begin{split}L_1 = \{ \{ (1, \text{hakuna}) \}, \\ \{ (1, \text{matata}) \} \}\end{split}\]

\[\begin{split}L_2 = \{ \{ (1, \text{hakuna}), (2, \text{hakuna}) \}, \\ \{ (1, \text{hakuna}), (2, \text{matata}) \}, \\ \{ (1, \text{matata}), (2, \text{hakuna}) \}, \\ \{ (1, \text{matata}), (2, \text{matata}) \} \}\end{split}\]

\[\begin{split}L_3 = \{ \{ (1, \text{hakuna}), (2, \text{hakuna}), (3, \text{hakuna}) \}, \\ \{ (1, \text{hakuna}), (2, \text{hakuna}), (3, \text{matata}) \}, \\ \{ (1, \text{hakuna}), (2, \text{matata}), (3, \text{hakuna}) \}, \\ \{ (1, \text{hakuna}), (2, \text{matata}), (3, \text{matata}) \}, \\ \{ (1, \text{matata}), (2, \text{hakuna}), (3, \text{hakuna}) \}, \\ \{ (1, \text{matata}), (2, \text{hakuna}), (3, \text{matata}) \}, \\ \{ (1, \text{matata}), (2, \text{matata}), (3, \text{hakuna}) \}, \\ \{ (1, \text{matata}), (2, \text{matata}), (3, \text{matata}) \} \}\end{split}\]

\[\text{...}\]

The Dialect is the union of all delimited Phrases in all Lexicons of the Language,

\[D = \{ \text{hakuna}, \text{matata}, \text{hakuna hakuna}, \text{hakuna matata}, \text{matata hakuna}, \text{matata matata}, ... \}\]

∎

Theorem 1.3.4

forall s in D: nexists j in N_{l(s)}: (s[j+1] = sigma) land (s[j] = sigma)

Proof Let \(s \in D\). Then by definition of the Dialect, for some \(i \in mathbb{N}\) and \(p \in L_i\),

\[s = \Pi_{j=1}^{i} p(j)\]

Where,

\[p = (\alpha_1, \alpha_2, ... , \alpha_i)\]

With \(\alpha_j \in L\) for \(1 \leq j \leq i\)

The proof proceeds by induction on the Word Length of the Lexicon, i.e. \(i\).

Basis Assume \(i = 1\)

Then by the definition of Limitations,

\[s = \alpha_1\]

By Discovery Axiom,

\[\forall j \in N_{l(\alpha_1)}: \alpha_1[j] \neq \sigma\]

Therefore,

\[\nexists j \in N_{l(s)}: (s[j+1] = \sigma) \land (s[j] = \sigma)\]

Induction. Assume for all \(t \in D\) with \(p \in L_i\) and some fixed \(i\).

\[\nexists j \in N_{l(t)}: (t[j+1] = \sigma) \land (t[j] = \sigma)\]

where

\[t = \Pi_{j=1}^{i} \alpha_j\]

Consider \(s \in D\) where \(p \in L_{i+1}\). Then, by definition of the Dialect,

\[s = \Pi_{j=1}^{i+1} \alpha_j\]

By the Induction clause of the definition of Limitations,

\[s = (\Pi_{j=1}^{i} \alpha_j)(\sigma)(\alpha_{i+1})\]

But, by inductive hypothesis, the quantity \(\Pi_{j=1}^{i} \alpha_j\) does not have consecutive Delimiters. Moreover, by the Induction clause of Limitations, this term can be rewritten

\[\Pi_{j=1}^{i} \alpha_j = (\Pi_{j=1}^{i-1} \alpha_j)(\sigma)(\alpha_i)\]

to show it must end in a Word. Therefore, by the Discovery Axiom, the quantity,

\[(\Pi_{j=1}^{i} \alpha_j)(\sigma)(\alpha_i)\]

cannot contain consecutive Delimiters.

The induction is thus established. Summarizing and generalizing,

\[\forall s \in D: \nexists j \in N_{l(s)}: (s[j+1] = \sigma) \land (s[j] = \sigma)\]