User blog:P進大好きbot/New Side Nesting Notation

This is an English translation of my Japanese blog post submitted to a Japanese googological event. This is a notation based on the side nesting method in Japanese googology. It might include many typos and errors, e.g. undefined behaviours or obvious infinite loops. I appreciate feedbacks.

= Summary =

Faery「This time, I'm gonna create an ordinal notation associated to Veblen hierarchy for y'all!」

Unfortunately, this is a notation not associated to Veblen hierarchy. She tried to create an ordinal notation associated to Veblen hierarchy, but made mistakes in the comarison algorithm and the standardness algorithm. Does it well-founded? I do not know.

= Notation =

I define a recursive set \(T\) of formal strings consisting of \((\), \\), and \(+\) in the following recursive way: I put \(0 = \). I define a recursive subset \(PT \subset T\) in the following way: For example, \((000) \in PT\) while \((000) + (000) \in T \setminus PT\).
 * 1) \( \in T\).
 * 2) For any \((a,b,c) \in T^3\), \((abc) \in T\).
 * 3) For any \((a,b) \in (T \setminus \{\})^2\), \(a+b \in T\).
 * 1) \(0 \notin PT\).
 * 2) For any \((a,b,c) \in T^3\), \((abc) \in PT\).
 * 3) For any \((a,b) \in (T \setminus \{0\})^2\), \(a+b \notin PT\).

Faery「Variadic Veblen hierarchy smells too complicated, so it's pretty good to start from \(3\)-ary one!」

= Ordering =

I simultaneously define recursive \(2\)-ary relations \(s < t\) and \(s \leq t\) on \(T\) in the following recursive way: Then \(\leq\) forms a total ordering, but does not form a well-ordering.
 * 1) \(s \leq t\) is equivalent to \(s < t\) or \(s = t\).
 * 2) If \(s = 0\), then \(s < t\) is equivalent to \(t \neq 0\).
 * 3) If \(s \neq 0\) and \(t = 0\), then \(s < t\) does not hold.
 * 4) Suppose that there exist an \((a,b,c) \in T^3\) and a \((d,e,f) \in T^3\) such that \(s = (abc)\) and \(t = (def)\):
 * 5) If \(a = d\), then \(s < t\) is equivalent to that either one of the following holds:
 * 6) \(b < e\) and \(c \leq t\).
 * 7) \(b = e\) and \(c < f\).
 * 8) If \(a \neq d\), then \(s < t\) is equivalent to that either one of the following holds:
 * 9) \(a < d\) and \(c \leq t\).
 * 10) \(d < a\) and \(s < f\).
 * 11) If there exist an \((a,b,c) \in T^3\) and a \((d,e) \in PT \times (T \setminus \{0\})\) such that \(s = (abc)\) and \(t = d+e\), then \(s < t\) is equivalent to \(s \leq d\).
 * 12) If there exist an \((a,b) \in PT \times (T \setminus \{0\})\) and a \((d,e,f) \in T^3\) such that \(s = a+b\) and \(t = (def)\), then \(s < t\) is equivalent to the negation of \(t < s\).
 * 13) If there exist an \((a,b) \in PT \times (T \setminus \{0\})\) and \((d,e) \in PT \times (T \setminus \{0\})\) such that \(s = a+b\) and \(t = d+e\), then \(s < t\) is equivalent to that either one of the following holds:
 * 14) \(a < d\).
 * 15) \(a = d\) and \(b < e\).

'''Faery「Somehow weird... Did I make mistakes...? But I've no idea...」'''

= Standard Form =

I put \(1 = (000)\). I define a recursive map \begin{eqnarray*} T \times T & \to & T \\ (a,t) & \mapsto & (a00) \times (t) \end{eqnarray*} in the following recursive way: For example, \((100) \times (1+1) = (100) + (100)\).
 * 1) If \(t = 0\), then \((a00) \times (t) = 0\).
 * 2) Suppose that there exists a \((d,e,f) \in T^3\) such that \(t = (def)\).
 * 3) Suppose that \(t < (00(00(a00)+1))\).
 * 4) If \(t = 1\), then \((a00) \times (t) = (a00)\).
 * 5) If \(d = 0\), \(e = 0\), and \(s \neq 1\), then \((a00) \times (t) = (00(a00)+f)\).
 * 6) If \(d = 0\) and \(e \neq 0\), then \((a00) \times (t) = (00(a00)+t)\).
 * 7) If \(d \neq 0\), then \((a00) \times (t) = (00(a00)+t)\).
 * 8) If \((00(00(a00)+1)) \leq t\), then \((a00) \times (t) = t\).
 * 9) If there exists a \((d,e) \in (T \setminus \{0\}) \times PT\) such that \(t = d+e\), then \((a00) \times (t) = (a00) \times (d) + (a00) \times (e)\).

For each \(x \in T\), I define a recursive subset \(OT_x \subset T\) in the following recursive way: I put \(OT = \{x \in T \mid x \in OT_x\}\). I call an expression in \(OT\) a standard form expression. For example, \((100)\) and \((100) + (010)\) are standard form expressions, while \((010)\) is not. I expect that the restriction of \(\leq\) to \(OT\) forms a well-ordering.
 * 1) \(0 \in OT_x\).
 * 2) For any \((a,b,c) \in T^3\), \((abc) \in OT_x\) is equivalent to that all of the following hold:
 * 3) \((a,b,c) \in OT_x^3\).
 * 4) If \(a = 0\), then \((100) \times (b) \leq x\).
 * 5) If \(a \neq 0\), then \((a+100) \times (b) \leq x\).
 * 6) \(b \in OT_{(a+100) \times (b)}\).
 * 7) \(c < (abc)\).
 * 8) For any \((s,t) \in (T \setminus \{0\}) \times PT\), \(s+t \in OT_x\) is equivalent to that all of the following hold:
 * 9) \((s,t) \in OT_x^2\).
 * 10) If \(s \in PT\), then \(t \leq s\).
 * 11) If there exists an \((a,b) \in T \times PT\) such that \(s = a+b\), then \(t \leq b\).

'''Faery「The weird comparison perhaps works by restricting additions in standard form expressions!! I WIN!!」'''

= Fundamental Sequence =

For an \(s \in T\), I denote by \(L(s)\) the length of \(s\) as formal strings. I define a recursive map \begin{eqnarray*} [ \ ] \colon OT \times \mathbb{N} & \to & OT \\ (s,n) & \mapsto & s[n] \end{eqnarray*} in the following recursive way: By the definition, \(s \neq 0\) implies \(s[n] < s\).
 * 1) If \(s = 0\), then \(s[n] = 0\).
 * 2) If \(s \neq 0\), then \(s[n] = \max \{t \in OT \mid t < s \land L(t) < L(s) + 9n\}\).

Faery「Setting FSs is awfully tiresome, so I throw it away in this way!」

= Large Function =

I define a recursive map \begin{eqnarray*} (100)_{\bullet}(\bullet) \colon OT \times \mathbb{N} & \to & \mathbb{N} \\ (s,n) & \mapsto & (100)_s(n) \end{eqnarray*} in the following recursive way: If the restriction of \(\leq\) to \(OT\) is a well-ordering, then \((100)_{\bullet}(\bullet)\) is total.
 * 1) If \(s = 0\), then \((100)_s(n) = n+1\).
 * 2) If \(s \neq 0\), then \((100)_s(n) = \sum_{m=0}^{n} (100)_{s[m]}^m(m)\).

I define a recursive map \begin{eqnarray*} (100)_{\bullet} \colon \mathbb{N} & \to & OT \\ n & \mapsto & (100)_n \end{eqnarray*} in the following recursive way: The recursive map \((100)_{\bullet}\) gives a limit of the notation system \((OT,\leq)\).
 * 1) If \(n = 0\), then \((100)_n = (100)\).
 * 2) If \(n \neq 0\), then \((100)_n = ((100)_{n-1}00)\).

Faery「Finally, I've achieved \(\varphi(1,0,0,0) = \varphi(\varphi(\cdots \varphi(1,0,0) \cdots,0,0),0,0)\), haven't I!?」

I note that her analysis is wrong, because it is not an ordinal notation associated to Veblen hierarchy. I expect that the limit of this notation is an OCF-level if it actually works.

= Large Number =

I submitted the computable large number \((100)_{(100)_{(100)_{(100)}(100)}}(100)\) to the Japanese googological event.

Faery oO(Yeah...! Everyone enjoys my ordinal notation associated to Veblen hierarchy...! Zzz...)