# Technically Exists

## Theorem $\mu$-2

The mu factorial has the same growth rate as $f_\omega$ in the Wainer hierarchy in terms of dominance over translations.

$n\mu! \eqv f_\omega(n)$

### Lemma $\mu$-5

$(2 \uparrow^{n - 1} n) + n < 2 \uparrow^n n$ for all $n \in \N_0, n \ge 3$.

#### Proof

Case where $n = 3$:

\begin{align*} (2 \uparrow^{3 - 1} 3) + 3 &= 16 + 3 \\ &= 19 \\ &< 65,536 \\ &= 2 \uparrow^3 3 \\ \end{align*}

Cases where $n \ge 4$:

\begin{align*} (2 \uparrow^{n - 1} n) + n &< (2 \uparrow^{n - 1} n) + (2 \uparrow^{n - 1} n) &&\text{by the Knuth Arrow Successor Inequality} \\ &= 2 \cdot (2 \uparrow^{n - 1} n) \\ &< 4 \uparrow^{n - 1} n &&\text{by the Knuth Arrow Multiplicative Inequality} \\ &\le n \uparrow^{n - 1} n \\ &= n \uparrow^n 2 \\ &< 2 \uparrow^n n &&\text{by the Knuth Arrow Commutative Inequality} \\ \end{align*}

### Lemma $\mu$-6

$n\mu! < 2 \uparrow^{n - 1} n$ for all $n \in \N_0$.

#### Proof

Base cases:

\begin{align*} 0\mu! &= 0 \\ &< 2 \\ &= 2 \uparrow^{0 - 1} 0 \\ 1\mu! &= 1 \\ &< 2 \\ &= 2 \uparrow^{1 - 1} 1 \\ 2\mu! &= 2 \\ &< 4 \\ &= 2 \uparrow^{2 - 1} 2 \\ 3\mu! &= 9 \\ &< 16 \\ &= 2 \uparrow^{3 - 1} 3 \\ \end{align*}

Induction step for $n \ge 3$:

\begin{align*} (n + 1)\mu! &= (n + 1) \uparrow^{n - 1} (n\mu!) \\ &< (n + 1) \uparrow^{n - 1} (2 \uparrow^{n - 1} n) &&\text{by the induction hypothesis} \\ &\le (2 \uparrow^{n - 1} n) \uparrow^{n - 1} (2 \uparrow^{n - 1} n) &&\text{by the Knuth Arrow Successor Inequality} \\ &< 2 \uparrow^{n - 1} ((2 \uparrow^{n - 1} n) + n) &&\text{by the Knuth Arrow Theorem} \\ &< 2 \uparrow^{n - 1} (2 \uparrow^n n) &&\text{by Lemma } \mu\text{-5} \\ &= 2 \uparrow^n (n + 1) \\ \end{align*}

### Proposition $\mu$-4

$n\mu! < f_\omega(n)$ for all $n \in \N_0$.

#### Proof

Case where $n = 0$:

\begin{align*} 0\mu! &= 0 \\ &< 1 \\ &= f_\omega(0) \\ \end{align*}

Cases where $n \ge 1$:

\begin{align*} n\mu! &< 2 \uparrow^{n - 1} n &&\text{by Lemma } \mu\text{-6} \\ &\le f_n(n) &&\text{by Proposition } \omega\text{-0} \\ &= f_{\omega[n]}(n) \\ &= f_\omega(n) \\ \end{align*}

### Proposition $\mu$-5

$(n + 1)\mu! > f_\omega(n)$ for all $n \in \N_0, n \ge 2$.

#### Proof

Case where $n = 2$:

\begin{align*} (2 + 1)\mu! &= 9 \\ &> 8 \\ &= f_\omega(2) \\ \end{align*}

Cases where $n \ge 3$:

\begin{align*} (n + 1)\mu! &= (n + 1) \uparrow^{n - 1} (n\mu!) \\ &> (n + 1) \uparrow^{n - 1} (2 \cdot (n + 1)) &&\text{by Lemma } \mu\text{-0} \\ &> (n + 1) \uparrow^{n - 1} (2 \cdot n) \\ &> 2 \uparrow^{n - 1} (2 \cdot n) \\ &\ge 2 \cdot f_\omega(n) &&\text{by Theorem } \omega\text{-1} \\ &> f_\omega(n) \\ \end{align*}

### Proof of Theorem $\mu$-2

By Proposition $\mu$-4, it cannot be the case that $n\mu! \gf f_\omega(n)$. By Proposition $\mu$-5, it cannot be the case that $n\mu! \gs f_\omega(n)$. Therefore, $n\mu! \eqv f_\omega(n)$.

### Referenced results

Knuth Arrow Successor Inequality: Theorem 3.5 in Knuth’s iterated powers

Knuth Arrow Multiplicative Inequality: Lemma 3.7 in Knuth’s iterated powers

Knuth Arrow Commutative Inequality: Theorem 3.2 in Knuth’s iterated powers

Knuth Arrow Theorem: Theorem 3.1 in Knuth’s iterated powers

Proposition 𝜔-0

Lemma 𝜇-0

Theorem 𝜔-1