Show that first order Peano's axioms capture the natural numbers regarding satisfiablity
$begingroup$
Denote be $mathcal P_{MO}$ the set of the monadic second order axioms of Peano. Then as shown by Dedekind any two model are isomorphic to $mathbb N$. Hence for a monadic second order sentence we have
$$
mathbb N models varphi mbox{ iff } mathcal P_{MO} models varphi
$$
i.e., the natural numbers fullfil $varphi$ iff every model that fulfills $mathcal P_{MO}$ also fulfills $varphi$.
Let $mathcal P_{FO}$ be the first order Peano arithmetic where the monadic induction axiom is replaced by a first order induction scheme over arbitrary predicates.
Let $psi$ be some first order sentence, then how to show that $mathcal P_{FO} models psi$ implies $mathbb N models psi$? (the other direction is clear, so I am asking about the relation between satisfiability of the axioms vs satisfiability in the natural numbers).
Similar for other first order axiomatization like Robinson arithmetic? The problem is that as mentioned here, the first order theories are not categorical.
logic arithmetic first-order-logic axioms peano-axioms
edited Dec 5 '18 at 15:10
Bernard
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
$endgroup$
add a comment |
$begingroup$
Denote be $mathcal P_{MO}$ the set of the monadic second order axioms of Peano. Then as shown by Dedekind any two model are isomorphic to $mathbb N$. Hence for a monadic second order sentence we have
$$
mathbb N models varphi mbox{ iff } mathcal P_{MO} models varphi
$$
i.e., the natural numbers fullfil $varphi$ iff every model that fulfills $mathcal P_{MO}$ also fulfills $varphi$.
Let $mathcal P_{FO}$ be the first order Peano arithmetic where the monadic induction axiom is replaced by a first order induction scheme over arbitrary predicates.
Let $psi$ be some first order sentence, then how to show that $mathcal P_{FO} models psi$ implies $mathbb N models psi$? (the other direction is clear, so I am asking about the relation between satisfiability of the axioms vs satisfiability in the natural numbers).
Similar for other first order axiomatization like Robinson arithmetic? The problem is that as mentioned here, the first order theories are not categorical.
logic arithmetic first-order-logic axioms peano-axioms
edited Dec 5 '18 at 15:10
Bernard
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
$endgroup$
3
$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
3
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36
add a comment |
$begingroup$
Denote be $mathcal P_{MO}$ the set of the monadic second order axioms of Peano. Then as shown by Dedekind any two model are isomorphic to $mathbb N$. Hence for a monadic second order sentence we have
$$
mathbb N models varphi mbox{ iff } mathcal P_{MO} models varphi
$$
i.e., the natural numbers fullfil $varphi$ iff every model that fulfills $mathcal P_{MO}$ also fulfills $varphi$.
Let $mathcal P_{FO}$ be the first order Peano arithmetic where the monadic induction axiom is replaced by a first order induction scheme over arbitrary predicates.
Let $psi$ be some first order sentence, then how to show that $mathcal P_{FO} models psi$ implies $mathbb N models psi$? (the other direction is clear, so I am asking about the relation between satisfiability of the axioms vs satisfiability in the natural numbers).
Similar for other first order axiomatization like Robinson arithmetic? The problem is that as mentioned here, the first order theories are not categorical.
logic arithmetic first-order-logic axioms peano-axioms
edited Dec 5 '18 at 15:10
Bernard
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
$endgroup$
Denote be $mathcal P_{MO}$ the set of the monadic second order axioms of Peano. Then as shown by Dedekind any two model are isomorphic to $mathbb N$. Hence for a monadic second order sentence we have
$$
mathbb N models varphi mbox{ iff } mathcal P_{MO} models varphi
$$
i.e., the natural numbers fullfil $varphi$ iff every model that fulfills $mathcal P_{MO}$ also fulfills $varphi$.
Let $mathcal P_{FO}$ be the first order Peano arithmetic where the monadic induction axiom is replaced by a first order induction scheme over arbitrary predicates.
Let $psi$ be some first order sentence, then how to show that $mathcal P_{FO} models psi$ implies $mathbb N models psi$? (the other direction is clear, so I am asking about the relation between satisfiability of the axioms vs satisfiability in the natural numbers).
Similar for other first order axiomatization like Robinson arithmetic? The problem is that as mentioned here, the first order theories are not categorical.
logic arithmetic first-order-logic axioms peano-axioms
logic arithmetic first-order-logic axioms peano-axioms
edited Dec 5 '18 at 15:10
Bernard
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
edited Dec 5 '18 at 15:10
Bernard
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
edited Dec 5 '18 at 15:10
Bernard
122k740116
edited Dec 5 '18 at 15:10
Bernard
122k740116
edited Dec 5 '18 at 15:10
Bernard
122k740116
122k740116
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
asked Dec 5 '18 at 15:06
StefanHStefanH
8,14652366
8,14652366
3
$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
3
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36
add a comment |
3
$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
3
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36
3
3
$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
3
3
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36
add a comment |
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$begingroup$
"the other direction is clear" The other direction is false, since first-order PA isn't complete.
$endgroup$
– Noah Schweber
Dec 5 '18 at 15:12
$begingroup$
@NoahSchweber This is not the provability-relation which I would denote by $vdash$, the notation $Sigma models varphi$ means that every model that satisifies $Sigma$, also satisfies $varphi$, and as $mathbb N$ satisfies $Sigma = mathcal P_{FO}$ this gives the mentioned direction...
$endgroup$
– StefanH
Dec 5 '18 at 15:14
3
$begingroup$
@StefanH The completeness theorem for first-order logic says that the relations $models$ and $vdash$ (between first-order theories and first-order sentences) are the same. The fact that $mathbb{N}models mathcal{P}_{FO}$ means it's clear that $mathcal{P}_{FO}models psi$ implies $mathbb{N}models psi$, not the other way around. And the converse is false, as Noah said, exactly because $mathcal{P}_{FO}$ is incomplete.
$endgroup$
– Alex Kruckman
Dec 5 '18 at 15:31
$begingroup$
Got it! Would anyone like to write up a full anwer....
$endgroup$
– StefanH
Dec 5 '18 at 15:36