Section II: Structures

Inversion, while a key component of the apparatus necessary for understanding the dynamics of Palindromes, is not the only linguistic operation involved in the formation of Palindromes. The pure involutive property of Palindromes (e.g., \(\zeta = {\zeta}^{-1})\) only manifests in a rare class of Sentences known as Perfect Palindromes, to be defined shortly.

However, the vast majority of Palindromes in any language are not pure involutions. Instead, the operation of inversion usually degrades the semantic content of a Sentence by re-ordering the Delimiters, as seen in the following,

\[\zeta = \text{"now sir a war is won"}\]

\[{\zeta}^{-1} = \text{"now si raw a ris won"}\]

In order to properly understand the nature of a Palindrome, the system requires a method of quantifying the distribution of Delimiters in a Sentence and making claims about the structure of that distribution. Furthermore, it must have a method of removing the “impurities” in Delimiter distributions that are introduced through inversion.

Section II.I: Delimiter Count

Definition 2.1.1: Delimiter Count

Let \(s \in S\). The Delimiter Count of \(s\), denoted \(\Delta(s)\), is defined inductively using the following schema,

• Basis:

  • If \(\neg(\sigma \subset_s s)\), then \(\Delta(s) = 0\)

  • If \(s = \varepsilon\), then \(\Delta(s) = 0\)

  • If \(s = \sigma\), then \(\Delta(s) = 1\)

• Induction:

  • If \(s = (u)(\varsigma)(v)\) for any \(u,v \in S\) where \(\neg(\sigma \subets_s u)\), \(\Delta(s) = Delta(v) + 1\)

Example Let \(ᚠ = \text{"with up so floating many bells down"}\). Consider \(\Delta(ᚠ)\).

Note

The astute reader will note there are 6 Delimiters in \(ᚠ\).

Let \(u_1 = \text{"with"}\) and \(v_1 = \{"up so floating many bells down"}\). Then, \(ᚠ = (u_1)(\varsigma)(v_1)\). Note \(\neg(\sigma \subset_s u_1)\). By the Induction clause of the Delimiter Count,

\[\Delta(ᚠ) = \Delta(v_1) + 1\]

Let \(u_2 = \text{"up"}\) and \(v_2 = \{"so floating many bells down"}\). Then, \(v_1 = (u_2)(\varsigma)(v_2)\). Note \(\neg(\sigma \subset_s u_2)\). By the Induction clause,

\[\Delta(v_1) = \Delta(v_2) + 1\]

Let \(u_3 = \text{"so"}\) and \(v_3 = \{"floating many bells down"}\). Then, \(v_2 = (u_3)(\varsigma)(v_3)\). Note \(\neg(\sigma \subset_s u_3)\). By the Induction clause,

\[\Delta(v_2) = \Delta(v_3) + 1\]

Let \(u_4 = \text{"floating"}\) and \(v_4 = \{"many bells down"}\). Then, \(v_3 = (u_4)(\varsigma)(v_4)\). Note \(\neg(\sigma \subset_s u_4)\). By the Induction clause,

\[\Delta(v_3) = \Delta(v_4) + 1\]

Let \(u_5 = \text{"many"}\) and \(v_5 = \{"bells down"}\). Then, \(v_4 = (u_5)(\varsigma)(v_5)\). Note \(\neg(\sigma \subset_s u_5)\). By the Induction clause,

\[\Delta(v_4) = \Delta(v_5) + 1\]

Let \(u_6 = \text{"bells"}\) and \(v_6 = \{"down"}\). Then, \(v_5 = (u_6)(\varsigma)(v_6)\). Note \(\neg(\sigma \subset_s u_6)\). By the Induction clause,

\[\Delta(v_5) = \Delta(v_6) + 1\]

Note \(\neg(\sigma \subset_s v_6)\). Therefore, by the Basis clause,

\[\Delta(v_6) = 0\]

Putting the recursion together,

\[\Delta(ᚠ) = (((((0 + 1) + 1) + 1) + 1) + 1) + 1 = 6\]

∎

Theorem 2.1.1

The Word Length of a Sentence in the Corpus is equal to one more than its Delimiter Count.

\[\forall \zeta \in C: \Lambda(\zeta) = \Delta(\zeta) + 1\]

Proof Let \(\zeta \in C\). Let \(n = \Lambda(\zeta)\). By the Theorem 1.4.3,

\[\zeta = \Pi_{i=1}^{n} \zeta[[i]]\]

By the definition of Limitations,

\[= (\zeta[[1]])(\varsigma)(\zeta[[2]]) ... (\varsigma)(\zeta[[n]])\]

The proof proceeds by induction on Word Length.

Basis: Assume \(n = 1\). Then,

\[\zeta = \zeta[[1]]\]

Let \(m = l(\zeta[[1]])\). By the Discovery Axiom,

\[\forall i \in N_m: (\zeta[[1]])[i] \neq \sigma\]

By the definition of Containment,

\[\neg(\sigma \subset_s \zeta[[1]])\]

By the Basis clause of the definition of Delimiter Counts,

\[\Delta(\zeta) = 0\]

Therefore,

\[\Lambda(\zeta) = \Delta(\zeta) + 1\]

Induction: Let \(\Lambda(\zeta) = n\) for some \(n \geq 1\). Assume \(\Lambda(\zeta) = \Delta(\zeta) + 1\)

Let \(\xi \in C\) such that \(Lambda(\xi) = n + 1\). Therefore, \(\sigma \subset_s \xi\). By definition of Containment,

\[\exists u,v \xi = (u)(\varsigma)(v)\]

Choose \(u = \xi[[1]]\) so that \(\neg(\varsigma \subset_s u)\).

Section II.II: Reductions

Definition 2.2.1: σ-Reduction

Let \(s \in S\). The σ-Reduction of \(s\), denoted \(\varsigma(s)\), is defined inductively with the following schema,

• Basis:

  • \(\varsigma(\sigma) = \varespilon\)

  • If \(neg(\sigma \subset_s s)\), then \(\varsigma(s) = s\)

• Induction:

  • If \(s = (u)(\varsigma)(v)\) for any \(u, v \in S\) where \(\neg(\sigma \subset_s u)\), \(\varsigma(s) = \varsigma(uv)\)

\(\varsigma(s)\) is said to be reduced from, or a reduction of, \(s\).

Example \(ᚠ = \text{"the widening circles into nothing gone"}\). Consider \(\varsigma(ᚠ)\)

Let \(u_1 = \text{"the"}\) and \(v_1 = \text{"widening circles into nothing gone"}\). Let \(w_1 = (u_1)(v_1)\). Then, \(ᚠ = (u_1)(\varsigma)(v_1)\). By the Induction clause of σ-Reduction,

\[\varsigma(ᚠ) = \varsigma((u_1)(v_1)) = \varsigma(w_1)\]

Let \(u_2 = \text{"thewidening"}\) and \(v_2 = \text{"circles into nothing gone"}\). Let \(w_2 = (u_2)(v_2)\). Then \(w_1 = (u_2)(\varsigma)(v_2)\). By the Induction clause,

\[\varsigma(w_1) = \varsigma((u_2)(v_2)) = \varsigma(w_2)\]

Let \(u_3 = \text{"thewideningcircles"}\) and \(v_3 = \text{"into nothing gone"}\). Let \(w_3 = (u_3)(v_3)\). Then \(w_2 = (u_3)(\varsigma)(v_3)\). By the Induction clause,

\[\varsigma(w_2) = \varsigmna((u_3)(v_3)) = \varsigma(w_3)\]

Let \(u_4 = \text{"thewideningcirclesinto"}\) and \(v_4 = \text{"nothing gone"}\). Let \(w_4 = (u_4)(v_4)\). Then \(w_3 = (u_4)(\varsigma)(v_4)\). By the Induction clause,

\[\varsigma(w_3) = \varsigmna((u_4)(v_4)) = \varsigma(w_4)\]

Let \(u_5 = \text{"thewideningcirclesintonothing"}\) and \(v_5 = \text{"gone"}\). Let \(w_5 = (u_5)(v_5)\). Then \(w_4 = (u_5)(\varsigma)(v_5)\). By the Induction clause,

\[\varsigma(w_4) = \varsigmna((u_5)(v_5)) = \varsigma(w_5)\]

But \(neg(\sigma \subset_s w_5)\), therefore by the Basis clause,

\[\varsigma(w_5) = w_5\]

Working back up through the recursion, the original reduction is found,

\[\varsigma(ᚠ) = \text{"thewideningcirclesintonothinggone"}\]

∎

Elementary Properties

1. From the Basis clause of σ-Reduction and the definition of String Length,

\[l(\varsigma(\sigma)) = l(\varepsilon) = 0\]

∎

2. From the Basis clause of σ-Reduction and the Discovery Axiom,

\[\varsigma(\alpha) = \alpha\]

∎

3. From the Basis clause of σ-Reduction, the Discovery Axiom and the definition of String Length

\[l(\varsigma(\alpha)) = l(\alpha)\]

∎

4. From the Induction clause of σ-Reduction and the definition of String Length,

\[l(\varsigma(s)) \leq l(s)\]

∎

5. From the definition of Concatenation and the definition of σ-Reduction, all σ-Reductions are Strings,

\[\forall s \in S: \varsigma(s) \in S\]

∎

Theorems

Theorem 2.2.1

The inverse of a reduced String is the reduction of the inverse String.

\[\forall s \in S: (\varsigma(s))^{-1} = \varsigma(s^{-1})\]

Proof Let \(s \in S\). The proof proceeds by induction on the number of Delimiters in \(s\).

Basis Let \(\neg(\sigma \subset_s s)\); that is, assume there are no Delimiter in \(s\). By Theorem 1.2.5 and the fact \(\sigma^{-1} = \sigma\),

\[\neg(\sigma \subset_s s) \equiv \neg(\sigma subset_s s^{-1})\]

Consider \((\varsigma(s))^{-1}\). By the Basis clause of the Reduction definition,

\[\varsigma(s) = s\]

Therefore,

\[(\varsigma(s))^{-1} = s^{-1}\]

Consider \(\varsigma(s^{-1})\). By \(\neg (\sigma \subset_s s^{-1})\) and the Basis clause of the Reduction definition,

\[\varsigma(s^{-1}) = s^{-1}\]

Induction Let \(\sigma \subset_s s\), i.e. \(s\) has \(k\) Delimiters for some \(k \geq 1\). Assume \((\varsigma(s))^{-1} = \varsigma(s^{-1})\).

Let \(u \in S\) such that \(u\) has \(k+1\) Delimiters. Let \(u = (v)(\varsigma)(w)\), where \(\neg(\sigma \subset_s v)\)

Then, by the definition of Containment, for some \(u,v\), possibly Empty,

\[s = (u)(\sigma)(v)\]

By Theorem 1.2.3,

\[s^{-1} = (v^{-1})(\sigma^{-1})(u^{-1})\]

By the definition of String Inversion and the fact \(l(\sigma) = 1\), it follows \(\sigma^{-1} = \sigma\). Therefore,

\[s^{-1} = (v^{-1})(\sigma)(u^{-1})\]

Consider \(\varsigma(s^{-1})\). Apply the Induction clause of the Reduction definition,

\[\varsigma(s^{-1}) = \varsigma((v^{-1})(u^{-1}))\]

Consider \(\varsigma(s)\). By the Induction clause of the Reduction definition,

\[\varsigma(s) = \varsigma(uv)\]

By Theorem 1.2.3,

\[(\varsigma(s))^{-1} = (\varsigma(uv))^{-1}\]

Section II.III: Palindromes

TODO

Section II.IV: Summary

TODO