provability of a formula in a given theory
$begingroup$
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
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
$endgroup$
add a comment |
$begingroup$
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
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
$endgroup$
add a comment |
$begingroup$
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
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
$endgroup$
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 '18 at 18:01
DaidalosDaidalos
303
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
asked Nov 29 '18 at 18:01
DaidalosDaidalos
303
303
add a comment |
add a comment |
1 Answer
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$begingroup$
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 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
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 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
$endgroup$
add a comment |
$begingroup$
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 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
$endgroup$
add a comment |
$begingroup$
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 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
$endgroup$
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 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
answered Nov 29 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
answered Nov 29 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
answered Nov 29 '18 at 18:21
Noah SchweberNoah Schweber
124k10150287
124k10150287
add a comment |
add a comment |
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