How to intuitively understand BMS

NOTE: This blog post focuses more on the behaviour of BMS rather than its ruleset. Also, this blog post uses BM2.

Primitive Sequence System[]

Think of it like this:

The empty string is 0.
(0) is 1.
Concatenation of strings of the form (0)X (X can be empty) is addition.
(0)X represents ω^ys for some y.
However, to represent y, we cannot use (0)Xs, so we use (1)Xs instead. (if y = 1, X = (1); if y = 2, X = (1)(1); etc.)
Then, (n+1)Xs are to (n)Xs as (1)Xs are to (0)Xs.

This is basically an encoding of Cantor's normal form, and so the limit is ε₀.

Pair Sequence System[]

Definition: An n-row BMS sequence X is standard if there exists an x such that by evaluating (0,0,0,...,0)(1,1,1,...,1)[x] in (n+1)-row BMS (where f(n) = n), we can eventually get to X[x].

Property: In a standard n-row BMS sequence, appending an entry at the start of each element of the sequence enumerating the elements starting from 0 always produces a standard expression in (n+1)-row BMS - and, in fact, the limit of the sequences produced is the limit of (n+1)-row BMS.

Then, we do some analysis:

Empty string <- Empty string = 0
(0,0) <- (0) = 1
(0,0)(1,0) <- (0)(0) = 2
(0,0)(1,0)(2,0) <- (0)(0)(0) = 3
(0,0)(1,0)(2,0)(3,0) <- (0)(0)(0)(0) = 4
(0,0)(1,1) <- (0)(1) = ω
(0,0)(1,1)(2,0) <- (0)(1)(0) = ω+1
(0,0)(1,1)(2,0)(3,0) <- (0)(1)(0)(0) = ω+2
(0,0)(1,1)(2,0)(3,1) <- (0)(1)(0)(1) = ω2
(0,0)(1,1)(2,0)(3,1)(3,0)(4,1) <- (0)(1)(0)(1)(0)(1) = ω3
(0,0)(1,1)(2,1) <- (0)(1)(1) = ω²
(0,0)(1,1)(2,1)(3,0) <- (0)(1)(1)(0) = ω²+1
(0,0)(1,1)(2,1)(3,0)(4,1) <- (0)(1)(1)(0)(1) = ω²+ω
(0,0)(1,1)(2,1)(3,0)(4,1)(5,1) <- (0)(1)(1)(0)(1)(1) = ω²2
(0,0)(1,1)(2,1)(3,1) <- (0)(1)(1)(1) = ω³
(0,0)(1,1)(2,1)(3,1)(4,1) <- (0)(1)(1)(1)(1) = ω⁴
(0,0)(1,1)(2,2) <- (0)(1)(2) = ω^ω
(0,0)(1,1)(2,2)(3,1) <- (0)(1)(2)(1) = ω^ω+1
(0,0)(1,1)(2,2)(3,1)(4,2) <- (0)(1)(2)(1)(2) = ω^ω2
(0,0)(1,1)(2,2)(3,2) <- (0)(1)(2)(2) = ω^ω²
(0,0)(1,1)(2,2)(3,2)(4,1)(5,2)(6,2) <- (0)(1)(2)(2)(1)(2)(2) = ω^ω²2
(0,0)(1,1)(2,2)(3,2)(4,2) <- (0)(1)(2)(2)(2) = ω^ω³
(0,0)(1,1)(2,2)(3,3) <- (0)(1)(2)(3) = ω^{ω^ω}
(0,0)(1,1)(2,2)(3,3)(4,2)(5,3) <- (0)(1)(2)(3)(2)(3) = ω^{ω^ω2}
(0,0)(1,1)(2,2)(3,3)(4,3) <- (0)(1)(2)(3)(3) = ω^{ω^ω²}
(0,0)(1,1)(2,2)(3,3)(4,4) <- (0)(1)(2)(3)(4) = ω^{ω^{ω^ω}}
(0,0)(1,1)(2,2)(3,3)(4,4)(5,5) <- (0)(1)(2)(3)(4)(5) = ω^{ω^{ω^{ω^ω}}}

From this analysis, we can deduce the following:

Property: Let X be a standard BMS sequence. If the rightmost element in X is of the form (x,y) for y > 1, find the least sequence Y in X such that Y is of the form (z,y-1)Z for some z < x and Z that contains the (x,y) (and can even be the (x,y) itself) and has no elements to the right of it. Then, when solving X[n] for some n, [WIP]