provability of a formula in a given theory
Suppose $T$ is an arithmetically sound theory. If there exists a polynomial function $p$ such that $T$ proves for every $n$
$negvarphi(overline{p(n)})rightarrowvarphi(overline{n})$
is it true that for every $n$, $T$ proves $varphi(overline{n})$
logic
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Suppose $T$ is an arithmetically sound theory. If there exists a polynomial function $p$ such that $T$ proves for every $n$
$negvarphi(overline{p(n)})rightarrowvarphi(overline{n})$
is it true that for every $n$, $T$ proves $varphi(overline{n})$
logic
add a comment |
Suppose $T$ is an arithmetically sound theory. If there exists a polynomial function $p$ such that $T$ proves for every $n$
$negvarphi(overline{p(n)})rightarrowvarphi(overline{n})$
is it true that for every $n$, $T$ proves $varphi(overline{n})$
logic
Suppose $T$ is an arithmetically sound theory. If there exists a polynomial function $p$ such that $T$ proves for every $n$
$negvarphi(overline{p(n)})rightarrowvarphi(overline{n})$
is it true that for every $n$, $T$ proves $varphi(overline{n})$
logic
logic
asked Nov 29 at 18:01
Daidalos
303
303
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No, and this has nothing to do with logic: take $varphi(x)$ to be "$x$ is even" and $p(x)=x+1$. The statement $$forall x(negvarphi(p(x))rightarrowvarphi(x))wedge negvarphi(overline{1})$$ is - to nuke a mosquito - provable in the arithmetically sound theory PA.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, and this has nothing to do with logic: take $varphi(x)$ to be "$x$ is even" and $p(x)=x+1$. The statement $$forall x(negvarphi(p(x))rightarrowvarphi(x))wedge negvarphi(overline{1})$$ is - to nuke a mosquito - provable in the arithmetically sound theory PA.
add a comment |
No, and this has nothing to do with logic: take $varphi(x)$ to be "$x$ is even" and $p(x)=x+1$. The statement $$forall x(negvarphi(p(x))rightarrowvarphi(x))wedge negvarphi(overline{1})$$ is - to nuke a mosquito - provable in the arithmetically sound theory PA.
add a comment |
No, and this has nothing to do with logic: take $varphi(x)$ to be "$x$ is even" and $p(x)=x+1$. The statement $$forall x(negvarphi(p(x))rightarrowvarphi(x))wedge negvarphi(overline{1})$$ is - to nuke a mosquito - provable in the arithmetically sound theory PA.
No, and this has nothing to do with logic: take $varphi(x)$ to be "$x$ is even" and $p(x)=x+1$. The statement $$forall x(negvarphi(p(x))rightarrowvarphi(x))wedge negvarphi(overline{1})$$ is - to nuke a mosquito - provable in the arithmetically sound theory PA.
answered Nov 29 at 18:21
Noah Schweber
120k10146279
120k10146279
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