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Dropping array notation (DAN) is the seventh part of my array notation.

Let’s have an overview of EAN, pDAN and sDAN. In EAN, the grave accent works as the least 1-separator, and it drops down once until the expansion. In pDAN, the double comma works as the least 2-separator, and it drops down twice until the expansion. In sDAN, the triple comma works as the least 3-separator, and it drops down 3 times until the expansion.

Continue upward, we have the least 4-separator – the quadruple comma (,,,,), the least 5-separator – the quintuple comma (,,,,,), and so on. Generally, an m-separator drops down m times until the expansion. Here comes the definition of dropping array notation.

# Definition

Syntax:

- s(a,b) is an array where a and b are positive integers.
- s(#Am) is an array, where m is a positive integer, A is a separator, and # is a string such that s(#) is an array.

Definition of separators:

- The m-ple comma (,,…,) with m ≥ 2 are separators. (basic separators)
- {m} is a separator, where m is a positive integer.
- {#Am} is a separator, where m is a positive integer, A is a separator, and # is a string such that {#} is a separator.

Note:

- The comma (,) is a shorthand for {1}.
- The dot . = {1{1,,2}2}, the grave accent ` = {1,,2}, the double comma ,, = {1,,,2}, the triple comma ,,, = {1,,,,2}, and so on, the m-ple comma ,,…, = {1,,…,,2} with an (m+1)-ple comma inside.

## Rules and process

- Rule 1: (base rule – only 2 entries) s(a,b) = a^b
- Rule 2: (tailing rule – the last entry is 1) s(# A 1) = s(#) and {# A 1} = {#}
- Rule 3: (recursion rule – neither the 2nd nor 3rd entry is 1) s(a,b,c #) = s(a, s(a,b-1,c #) ,c-1 #)

where # is a string of entries and separators, it can also be empty. A is a separator.

If none of the 3 rules above applies, start the process shown below. Note that case B1, B2 and B4 are terminal but case A and B3 are not. b, n, m, M, B_{u}′s and K are parts of the original array but j, t, u(j), v(j), i, S_{i}, A_{u}′s, X, Y, P and Q are not. Before we start, let A_{0} be the whole array. First start from the 3rd entry.

- Case A: If the entry is 1, then you jump to the next entry.
- Case B: If the entry n is not 1, look to your left:
- Case B1: If the comma is immediately before you, then
- Change the “1,n” into “b,n-1” where n is this non-1 entry and the b is the iterator.
- Change all the entries at base layer before them into the base.
- The process ends.

- Case B2: If the m-ple comma (m ≥ 2) M is immediately before you, then
- Change “M n” into “M 2 M n-1”, and let B
_{u(m+1)}be the former M. - Let t be such that the M is at layer t. And let B
_{0}be the whole array now. - Repeat this:
- Subtract t by 1.
- Let separator B
_{t}be such that it’s at layer t, and the M is inside it. - If t = 1, then break the repeating, or else continue repeating.

- Find the maximum of u(m) such that lv(A
_{u(m)}) < lv(M). - If lv(A
_{u(m)}) < lv(,), then- Let string P and Q be such that B
_{u(m)}= “P B_{u(m)+1}Q”. - Change B
_{u(m)}into “P {1 A_{u(m)+1}2} Q”. - The process ends.

- Let string P and Q be such that B
- Set j = m. Set A
_{u(m+1)}= M. - Repeat this:
- Subtract j by 1.
- Find the maximum of u(j) such that lv(A
_{u(j)}) < lv(A_{u(j+1)}). - Let string X and Y be such that B
_{u(j+1)}= “X B_{u(j+2)}2 Y”. - If lv(A
_{u(j)}) < lv(“X Y”), then- Find the minimum of v(j) such that v(j) > u(j) and lv(A
_{v(j)}) < lv(A_{u(j+2)}). - Let string P and Q be such that B
_{u(j)}= “P B_{v(j)}Q”. - Change B
_{u(j)}into “P X A_{v(j)}2 Y Q”. - The process ends.

- Find the minimum of v(j) such that v(j) > u(j) and lv(A
- If j = 1, then break the repeating, or else continue repeating.

- Let string P and Q be such that B
_{u(1)}= “P B_{u(2)}Q”. - Change B
_{u(1)}into S_{b}, where b is the iterator, S_{1}is comma, and S_{i+1}= “P S_{i}Q”. - The process ends.

- Change “M n” into “M 2 M n-1”, and let B
- Case B3: If a separator K is immediately before you, and it doesn’t fit case B1 or B2, then
- Change the “K n” into “K 2 K n-1”.
- Set separator A
_{t}= K, here K is at layer t. - Jump to the first entry of the former K.

- Case B4: If an lbrace is immediately before you, then
- Change separator {n #} into string S
_{b}, where b is the iterator, S_{1}= “{n-1 #}” and S_{i+1}= “S_{i}1 {n-1 #}”. - The process ends.

- Change separator {n #} into string S

- Case B1: If the comma is immediately before you, then

## Level comparison

All arrays have the lowest and the same level. And note that the same separators have the same level. To compare levels of other separators A and B, we follow these steps.

First some preparation.

- Find the maximum of m that there’re some m-ple commas in the expressions of A or B.
- Convert all the multiple comma of A and B into expressions of m-ple comma, using ,,…, (i commas) = {1,,…,,2} (i+1 commas)

Then the m-ple comma has the highest level, and other separators have level lower than it.

- Apply rule 2 to A and B until rule 2 cannot apply any more.
- Let A = {a
_{1}A_{1}a_{2}A_{2}…a_{k-1}A_{k-1}a_{k}} and B = {b_{1}B_{1}b_{2}B_{2}…b_{l-1}B_{l-1}b_{l}} - If k = 1 and l > 1, then lv(A) < lv(B); if k > 1 and l = 1, then lv(A) > lv(B); if k = l = 1, follow step 4; if k > 1 and l > 1, follow step 5 ~ 10
- If a
_{1}< b_{1}, then lv(A) < lv(B); if a_{1}> b_{1}, then lv(A) > lv(B); if a_{1}= b_{1}, then lv(A) = lv(B) - Let , and .
- If lv(A
_{maxM(A)}) < lv(B_{maxM(B)}), then lv(A) < lv(B); if lv(A_{maxM(A)}) > lv(B_{maxM(B)}), then lv(A) > lv(B); or else – - If |M(A)| < |M(B)|, then lv(A) < lv(B); if |M(A)| > |M(B)|, then lv(A) > lv(B); or else –
- Let A = {#
_{1}A_{maxM(A)}#_{2}} and B = {#_{3}B_{maxM(B)}#_{4}} - If lv({#
_{2}}) < lv({#_{4}}), then lv(A) < lv(B); if lv({#_{2}}) > lv({#_{4}}), then lv(A) > lv(B); or else – - If lv({#
_{1}}) < lv({#_{3}}), then lv(A) < lv(B); if lv({#_{1}}) > lv({#_{3}}), then lv(A) > lv(B); if lv({#_{1}}) = lv({#_{3}}), then lv(A) = lv(B)

# Explanation

The A_{t} in case B3 and the B_{t} in step 1 ~ 3 of case B2 are preparation. Step 4 and 5 are where the m-ple comma (the least m-separator) drops down to an (m-1)-separator. Step 7 is where the (m-1)-separator drops down m-1 times, to a 0-separator, and step 6 is a preparation for it. And step 8 ~ 10 are the expansion. Here, the A_{u(m+1)} = M is an m-separator, and the A_{u(j)} is a (j-1)-separator.

The step 5 and step 7d are where the adding rules apply. Once an adding rule apply, the expansion doesn’t apply; only when all the adding rules don’t apply, the expansion applies.

The level comparison rules also change. The preparation step change A and B into such that the m-ple comma is the only “basic separator”.

,, corresponds to M.

,,, corresponds to K.

,,,, corresponds to Pi^1_2.

So the limit ordinal is psi(Pi^1_w).

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