[saithound]

Your argument is wrong. The function that encodes every Integer as an element of Cantor, the one you call `p`, has signature `p: Integer -> Cantor`. This means that the compositions `(f.p)` and `(g.p)` are not defined. The compositions in the opposite order, `(p.f)` and `(p.g)`, are defined, but have signature `Integer->Cantor`, which is not in the form `Cantor->y`.

Your comment, "the post gets around this by only looking at the first n elements of every binary sequence" is also wrong. This is not at all what the `equal` function does.

[anfelor]

You are right, a decoding function is needed. The decoding for my given encoding function is not computable because you need to look at all elements of the sequence to see if the rest are the padded zeros or part of the number (because there is a '1' somewhere). But here is a computable, surjective decoding function: 1. The first bit tells you whether the integer should be positive or negative. 2. Then we parse blocks of ten bits. The first bit tells you whether this block is part of the number ('1') or if we reached the end ('0'). If it is part of the number, then the next nine bits are part of the binary encoding of the integer.

This will not terminate for some sequences (for example the all '1' sequence), but it terminates for all sequences that denote integers. Wouldn't this encoding be valid? (I don't know too much about the halting problem so I might well be wrong here).

> This is not at all what the `equal` function does.

Could you explain what it does do? The way I see it the `find` function constructs a counter-example lazily. The `p` predicate is then applied to this counter-example in `forsome`: `forsome p = p(find(\a -> p a))`. Only if the `p` predicate can determine if the counter-example is valid by looking at a finite amount of digits will the function terminate.

[tetromino_]

Let's try fixing grandparent's argument: let p be the decoder function, Cantor -> Integer. We can define it as the length of the initial run of zeroes (treat the initial bit as sign). This almost works.

Where it fails is the infinite sequence of zeroes: p would count forever; f.p and g.p are not decidable.

So grandparent's argument in fact shows that OpenCantor -> y does not have decidable equality (where OpenCantor is Cantor with the zero sequence - or by extension, any one computable element - excluded). One special element makes all the difference!