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I am trying to understand the proof of the aforementioned fact. For context, here's a rough outline of [what I am understanding to be] the idea behind the proof: in the language we construct a formula $\phi(x)$ asserting "machine $x$, when started on input $x$, halts in less than $f(|x|)$ steps", where $f:x\mapsto 2^{2^{kx}}$ for some $k$. Next, given any machine $M$ intending to decide $PrA$, we construct a machine $M'$ as follows: when started on input $w$, $M'$ overwrites it with $\neg\phi(w)$, then runs $M$ to decide it. If $M$ decides $\neg\phi(w) \in PrA$, $M'$ halts, otherwise it loops forever. The conclusion is that any machine that decides $PrA$ must take $\geq f(|\sharp M'|)$ steps to decide $\neg\phi(\sharp M')$ - otherwise we get a contradiction when we run $M'$ on $\sharp M'$. So far so good, I guess.

I get lost when it is time to construct $\phi$, though. For example, one of the building blocks [used in the construction] of $\phi$ is a formula $I_{n}(x)$, $|I_{n}(x)| = O(n)$, asserting "$x$ is less than $f(n)^{2}$". Now, the paper doesn't show how to construct this formula, and I have no idea how to, too, given that we can only use addition and we have that bound there.

What am I missing?

EDIT, as suggested: the proof I am talking about is in Rabin and Fischer's paper - .pdf available here.1 And one further thing: I found one post from Lipton's blog where he talks about the "trick" used to construct $I_{n}$ succinctly. It is a sketch only and unfortunately not enough for me. It is mentioned however that the trick was previously discovered by Fischer and Meyer, and Volker Strassen, independently, but I couldn't track down these papers describing it in detail - if anyone knows about them, please tell me!

1Michael J. Fischer, Michael O. Rabin: Super-exponential complexity of Presburger arithmetic. Also: CiteSeerX, MIT, DOI: 10.1007/978-3-7091-9459-1_5, http://hcs64.com/files/cc.pdf#page=35

Martin Sleziak
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wet
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    It would probably help to reference the exact source you're reading the proof in, such that answerers have the opportunity to try to align the structure and notation of their answers to it. – hmakholm left over Monica Oct 20 '16 at 21:34

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Fischer and Rabin don't explicitly construct $_()$. Instead they claim its existence is clear in the proof of Theorem 11. This is essentially true, since the multiplication formula $M_n(x,y,z)$ they construct in the proof of Theorem 8 satisfies $M_n(x,y,z) \leftrightarrow x < 2^{2^n} ∧ x·y = z ∧ x ∈ N.$

This fact can be used to define $_()$. Namely, set $_() = M_{n+1}(x,0,0)$ for x ∈ N. Since x·0 = 0 and x ∈ N, $M_{n+1}(x,0,0) \leftrightarrow x < 2^{2^{n+1}} = (2^{2^n})^2$

Rick
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