User blog comment:Rgetar/Three properties of standard forms/@comment-35470197-20181124090716/@comment-32213734-20181124183652

I didn't know Property 3 until about 2 week ago. Since August 2018 I tried to create a program, calculating ordinals, but it didn't work beyond BHO because of this property. Recently I carefully investigated it, found this property and finally make the program work.

"compute x[n]": here x is already given in some form, and we already have algorithm of calculating fs element. So, we get only one fs of x.

"compute f(x[n])": generally it means just substitute x[n] into f(x) instead of x, that is replace x with x[n] in f(x), but I'm not completely sure.

I didn't give an example, since this program uses my OCF "Veblen-like collapsing function", but I have not published it yet. So, example:

φ1(Ω2Ω) = φ1(Ω2φ(φ1(Ω2Ω) + 1) + φ1(Ω2Ω))

Here the right part is non-standard form of left part, which is standard form. (φ(X) is function with countable output for all X, and φ1(X) is function with uncountable output of cardinality Ω for all X).

But despite this

φ(φ1(Ω2Ω)) < φ(φ1(Ω2φ(φ1(Ω2Ω) + 1) + φ1(Ω2Ω)))

since right part contains countable ordinal φ(φ1(Ω2Ω) + 1), which is larger than left part, and φ(X) cannot be less than φ(Y) expression inside X (the same way as extended Veblen function cannot be less than any of its arguments).

So, let we compute φ(α)[0], and for this α should be

φ(α)[n] = φ(α[n])

Let

α[0] = φ1(Ω2φ(φ1(Ω2Ω) + 1) + φ1(Ω2Ω))

which is non-standard form of φ1(Ω2Ω). But we should not convert it into standard form. We just substitute it into φ(α):

φ(α) → φ(φ1(Ω2φ(φ1(Ω2Ω) + 1) + φ1(Ω2Ω)))

And only now we convert the whole expression into standard form (but in this case it is already standard form).

Otherwise, if we convert φ1(Ω2φ(φ1(Ω2Ω) + 1) + φ1(Ω2Ω)), then expression "φ(φ1(Ω2Ω) + 1)" will be lost, but it is this expression that makes the result so large, and we get much lesser ordinal φ(φ1(Ω2Ω)) instead, and this is error.