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fixed a set I had wrong
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user5810
user5810

Let $n$ be a member of $\omega$, which the omnipotent provers $Y$ and $N$ know. A Turing machine will be run starting with the $n$ already inputted, and the machine can have a natural number inputted from the prover of its choice, with no limit on the number of times it can do so.


Let $S$ be a subset of $\omega$.

Definition: S is decidable from competing provers if and only if

there exists a Turing machine such that, for all members $n$ of $\omega$, when run as above,
(1) if $n\in S$ then $Y$ has a strategy that will force the Turing machine to accept
(0) else $N$ has a strategy that will force the Turing machine to reject



What subsets $S$ of $2^{\omega}$$\omega$ are decidable from competing provers?
What if the Turing machine is required to always halt?
In each case, is the resulting class low for itself?

Let $n$ be a member of $\omega$, which the omnipotent provers $Y$ and $N$ know. A Turing machine will be run starting with the $n$ already inputted, and the machine can have a natural number inputted from the prover of its choice, with no limit on the number of times it can do so.


Let $S$ be a subset of $\omega$.

Definition: S is decidable from competing provers if and only if

there exists a Turing machine such that, for all members $n$ of $\omega$, when run as above,
(1) if $n\in S$ then $Y$ has a strategy that will force the Turing machine to accept
(0) else $N$ has a strategy that will force the Turing machine to reject



What subsets $S$ of $2^{\omega}$ are decidable from competing provers?
What if the Turing machine is required to always halt?
In each case, is the resulting class low for itself?

Let $n$ be a member of $\omega$, which the omnipotent provers $Y$ and $N$ know. A Turing machine will be run starting with the $n$ already inputted, and the machine can have a natural number inputted from the prover of its choice, with no limit on the number of times it can do so.


Let $S$ be a subset of $\omega$.

Definition: S is decidable from competing provers if and only if

there exists a Turing machine such that, for all members $n$ of $\omega$, when run as above,
(1) if $n\in S$ then $Y$ has a strategy that will force the Turing machine to accept
(0) else $N$ has a strategy that will force the Turing machine to reject



What subsets $S$ of $\omega$ are decidable from competing provers?
What if the Turing machine is required to always halt?
In each case, is the resulting class low for itself?

Source Link
user5810
user5810

What sets are "decidable from competing provers"?

Let $n$ be a member of $\omega$, which the omnipotent provers $Y$ and $N$ know. A Turing machine will be run starting with the $n$ already inputted, and the machine can have a natural number inputted from the prover of its choice, with no limit on the number of times it can do so.


Let $S$ be a subset of $\omega$.

Definition: S is decidable from competing provers if and only if

there exists a Turing machine such that, for all members $n$ of $\omega$, when run as above,
(1) if $n\in S$ then $Y$ has a strategy that will force the Turing machine to accept
(0) else $N$ has a strategy that will force the Turing machine to reject



What subsets $S$ of $2^{\omega}$ are decidable from competing provers?
What if the Turing machine is required to always halt?
In each case, is the resulting class low for itself?