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Let $\phi_1$ and $\phi_2$ be the following statements:

$\phi_1:$ There is a function $f:\{0,1\}^*\to\{0,1\}$ computable in $E$ that has circuit complexity $2^{\Omega(n)}$.

$\phi_2:$ There is a function $f:\{0,1\}^*\to\{0,1\}$ computable in $NE \cap CoNE$ that has $2^{\Omega(n)}$ hardness on average.

Q1. Is there any $\Pi_2$ sentence $\psi$ in the language of arithmetic such that we can prove $\mathbb{N}\models \psi \leftrightarrow \phi_1$?

Q2. Is there any $\Pi_2$ sentence $\psi$ in the language of arithmetic such that such that we can prove $\mathbb{N}\models \psi \leftrightarrow \phi_2$?

Actually, I want to know if these conjectures are $\Pi_2$ expressible like $P\not = NP$.

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    $\begingroup$ In $\mathbb N$? Of course. One of $\bot$ or $\top$ can be taken as $\phi_i$. $\endgroup$ Commented Aug 24, 2017 at 18:51
  • $\begingroup$ @EmilJeřábek: I edited my question. I need an explicit sentence $\psi$ such that we can prove it is equivalent to $\phi_1$ for example. $\endgroup$ Commented Aug 24, 2017 at 19:34
  • $\begingroup$ @EmilJeřábek Are you sure $\phi_i$ is provable in $\mathbb N$? $\endgroup$ Commented Aug 25, 2017 at 7:32
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    $\begingroup$ @FanZheng $\mathbb N$ is a model, not a theory. $\endgroup$ Commented Aug 25, 2017 at 7:59
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    $\begingroup$ That statement is not $\Pi_2$, but $\Sigma_2$ (just as the original expression): $$\exists n_0\exists \epsilon>0\,\forall n>n_0\text{$L_n$ requires circuits of size $\ge2^{\epsilon n}$}.$$ The two existential quantifiers are extremely important. $\endgroup$ Commented Aug 25, 2017 at 9:57

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As given, $\phi_1$ and $\phi_2$ are $\Sigma_2$.

They cannot be shown equivalent to $\Pi_2$ statements by any proof that relativizes. This follows by the same argument as in Examples of $G_\delta$ sets or https://cstheory.stackexchange.com/a/16644. We only need to know that the relativized statements are unaffected by a finite change of the oracle (which is obvious), and that either statement can be made true or false by relativization with suitable oracles:

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  • $\begingroup$ Why the following isn't equivalent to the $\phi_1$? $$\forall c \forall \epsilon(0<\epsilon<1 \to \exists n\forall C(size(C)\leq 2^{cn^\epsilon}\to \text{$C$ does not compute $L_n$}))$$ for a complete language $L\in E$? $\endgroup$ Commented Aug 25, 2017 at 15:46
  • $\begingroup$ Imagine, for example, that the circuit complexity of $L_n$ is $2^{n/\log n}$. Then $\phi_1$ is false, but your formula is true. $\endgroup$ Commented Aug 25, 2017 at 15:51
  • $\begingroup$ does this mean that assuming $size_L(n)$ dominates all elementary functions that are dominated by every $2^{cn}$ does not imply $\phi_1$? $\endgroup$ Commented Aug 25, 2017 at 15:59
  • $\begingroup$ Also, is $$\forall c \forall \epsilon(0<\epsilon<1 \to \exists n\forall C(size(C)\leq 2^{cn^\epsilon}\to \text{$C$ does not compute $L_n$}))$$ sufficient for derandomizing $BPP$? $\endgroup$ Commented Aug 25, 2017 at 16:01
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    $\begingroup$ On second thought, what I wrote earlier is not right. “$s_L(n)$ dominates all elementary functions that are dominated by every $2^{cn}$” is weaker than $\phi_1$. The correct equivalent is “$s_L(n)$ dominates all elementary functions that are infinitely often below any $2^{cn}$”. That is, $\forall f((\forall c>0\forall n_0\exists n\ge n_0\,f(n)\le2^{cn})\to\exists n_0\forall n\ge n_0\,s_L(n)\ge f(n))$. Yes, it is also equivalent if the last part is replaced with $\exists n_0\,s_L(n_0)\ge f(n_0)$. Either way, it is $\Pi_3$. $\endgroup$ Commented Aug 25, 2017 at 18:12

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