User blog comment:Tetramur/Pentational arrays and beyond - comparisons/@comment-37993808-20200108154849/@comment-39541634-20200109083113

I currently can't find Sbiis Saibian's explanation anywhere. If anyone can provide a source I'll be most grateful.

Going from memory, though:

In Saibian's method, X^^(X+1) does not transform into p^^(p+1). You are not allowed to change the X's to p's at that stage.

Instead, we visualize X^^(X+1) as a tower of height w+1: X^X^X^... with an additional "^X" appended at the top. He then defines this tower as some kind of limit, using a clever approach that I don't remember right now. I *do* remember that this approach involves climbing up the infinite tower of X's.

It is only after Saibian did all this preliminary work, that he is ready to replace the top X with a p. 

This works well, because Saibian's method closely mimics the usual way we create fundamental sequences: We only "expand" one transfinite symbol, keeping the rest of the structure intact.

In BEAF, however, the expression on the left of the "&" must give you the precise number of elements in the array. We can still organize these elements in any structure we want, but their number is fixed. Moreover, we are never ever allowed to increase the "geometrical" scope of our structure. As these structures represent ordinals, they must form a decending sequence as we expand them.

Note that the structure can become larger in size. If we start with 3^3 & 3, which is a 3-by-3 cube , the array will quickly grow into an immense size in two of the three dimensions. But it will never escape the confines of 3D. It won't even escape the confines three 2D grids. That is what I mean by "a decendng sequence".

And I don't see how these requirements could be compatible with Saibian's approach.

Of-course, there's a far simpler way to show the problem by comparing the number of elements of two allegedly equivalent structures. Let's pick a random such pair from Tetramur's blog post:

(X^^X)^(X*3) = X^X^(X^^X+3)

Setting X to (say) 2 we get:

(2^^2)^(2*3) = 2^2^(2^^2+3)

Or

4096 = 2^128

So the left array has 4096 elements while the right array has 2^128 elements. They cannot be the same array, regardless of how the numbers are arranged in it. Case closed.

(Of-course it is concievable that the two arrays would evaluate to the same number, but that's not what the claimed equality stated. It's an equality of structures, rather than an equality of numbers)

P.S.

You might wonder why these problems don't crop up before we reach tetrational arrays. The reason is simple: Below epsilon-0, there's a direct correspondence between BEAF expressions and Cantor Normal Forms (where X plays the role of omega).

This correspondence breaks down at tetrational arrays. We have (X^^X) that's supposed to correspond to e0, but (X^^X)^(X^^X) does not correspond to e0^e0 or anything like it. The firm connection to ordinal arithemtic is broken, which gets you all kind of strange results.