Appendix I: Addendums

TODO

Appendix I.I: Omitted Axioms

This section of the Appendix contains Axioms the author considered at various points during the formalization, but as they did not ultimately seem necessary to establish the main results of the work, they have been placed here.

Axiom S.2: Duality Axiom I

For every Sentence in the Corpus, there exists a Word in the Language which is contained in it.

\[\forall \zeta \in C: \exists \alpha \in L: \alpha \subset_s \zeta\]

Axiom S.3: Duality Axiom II

For every Word in the Language, there exists a Sentence in the Corpus that contains it.

\[\forall \alpha \in L: \exists \zeta \in C: \alpha \subset_s \zeta\]

Note

The Duality Axioms are reminiscent of the relation of surjectivity in real analysis. However, containment is not a strict equality relation so this resemblance should not be taken too far.

Axiom S.4: Density Axiom

There exists a natural number such that for all numbers less than it there exists a Sentence in the Corpus with that Word Length.

exists n in mathbb{N}: forall i in N_n: exists zeta in C: Lambda(zeta) = i

This Axiom is required to induce proofs about Word Length. Without it, there is no formal way to accomplish an induction over Word Length. However, it may be (justifiably) argued there is no reason a natural language should obey this rule. It is possible to conceive of a natural language that does not have sentences with, say, exactly \(n = 13\) words, perhaps due to an eccentricity in its grammar. In natural languages such as these, the results that depend on the following theorem are only valid in a truncated Corpus where all sentences with \(\Lambda(\zeta) > 12\) are ignored.

Appendix I.II: Omitted Proofs

Section I

Theorem 1.2.1

The String Length of the concatenation of String \(s\) and String \(t\) is equal to the sum of their String Lengths.

\[\forall s,t \in S: l(st) = l(s) + l(t)\]

Proof The proof proceeds by induction on \(t\).

Basis: Let \(t = \varepsilon\) and \(s \in S\). Consider \(st = s\varepsilon\).

By the basis clause of concatenation, \(s\varepsilon = s\). By the basis clause of String Length, \(l(\varepsilon) = 0\). It follows from the basic laws of arithmetic,

\[l(s\varepsilon) = l(s) = l(s) + 0\]

\[= l(s) + l(\varepsilon) = l(s) + l(t)\]

Therefore, the base case, \(l(st) = l(s) + l(t)\), holds.

Induction: Let \(s, t \in S\) and u in Sigma_{e}. Assume \(l(st) = l(s) + l(t)\). Let \(v = tu\) and consider,

\[l(sv) = l(s(tu)) = l((st)u)\]

If \(u = \varepsilon\), then applying the argument of the base case,

\[l(sv) = l((st)u) = l(st) + l(\varepsilon)\]

\[= l(st) = l(s) + l(t)\]

Where the last equality follows from the inductive hypothesis. Note \(t = t\varepsilon = tu = v\) by the basis clause of concatenation. From this, it follows the inductive step is established for \(u = \varepsilon\),

\[l(sv) = l(s) + l(v)\]

If \(u \neq \varepsilon\), then it follows from the induction clause of String Length,

\[l((st)u) = l(st) + 1 = l(s) + l(t) + 1 \quad \text{ (1) }\]

Where the last equality follows from the inductive hypothesis. Consider the quantity \(l(tu)\). By the induction clause of String Length,

\[l(tu) = l(t) + 1\]

Adding \(l(s)\) to both sides,

\[l(s) + l(tu) = l(s) + l(t) + 1 \quad \text{ (2) }\]

Comparing the RHS of (1) and (2), it follows the LHS are equal,

\[l(stu) = l(s) + l(tu)\]

Summarizing, if \(l(st) = l(s) + l(t)\) and \(u \in \Sigma_{e}\), then \(l(stu) = l(s) + l(tu)\). Therefore, the inductive step is established.

Since the basis case and inductive step have both been established, it follows from the principle of finite induction,

\[\forall s,t \in S: l(st) = l(s) + l(t)\]

∎

Theorem 1.2.2

The Empty Character is contained in every String.

\[\forall s \in S: \varepsilon \subset_s s\]

Proof Let \(s \in S\). By the Basis clause of Concatenation,

\[\varepsilon = \varepsilon\varepsilon\]

Therefore,

\[s = {\varepsilon}s = {\varepsilon\varepsilon}s\]

Let \(w_1 = \varepsilon\) and \(w_2 = s\). Then, \(s = {w_1}\varepsilon{w_2}\). By the definition of Containment,

\[\varepsilon \subset_s s\]

∎