A Theory Entailing a Logical Equivalence of Two Formulas in FOL
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In FOL we say two formulas $phi$ and $psi$ are logically equivalent iff $phivDashpsi$ and $psivDashphi$.
My question is: What does it mean to say that a theory $T$ entails that $phi$ and $psi$ are logically equivalent? $T vDash (phivDashpsi text{ and } psivDashphi)$
Can we say it means $T, phivDashpsi$ and $T, psivDashphi$?
logic first-order-logic
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
$endgroup$
add a comment |
$begingroup$
In FOL we say two formulas $phi$ and $psi$ are logically equivalent iff $phivDashpsi$ and $psivDashphi$.
My question is: What does it mean to say that a theory $T$ entails that $phi$ and $psi$ are logically equivalent? $T vDash (phivDashpsi text{ and } psivDashphi)$
Can we say it means $T, phivDashpsi$ and $T, psivDashphi$?
logic first-order-logic
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
$endgroup$
add a comment |
$begingroup$
In FOL we say two formulas $phi$ and $psi$ are logically equivalent iff $phivDashpsi$ and $psivDashphi$.
My question is: What does it mean to say that a theory $T$ entails that $phi$ and $psi$ are logically equivalent? $T vDash (phivDashpsi text{ and } psivDashphi)$
Can we say it means $T, phivDashpsi$ and $T, psivDashphi$?
logic first-order-logic
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
$endgroup$
In FOL we say two formulas $phi$ and $psi$ are logically equivalent iff $phivDashpsi$ and $psivDashphi$.
My question is: What does it mean to say that a theory $T$ entails that $phi$ and $psi$ are logically equivalent? $T vDash (phivDashpsi text{ and } psivDashphi)$
Can we say it means $T, phivDashpsi$ and $T, psivDashphi$?
logic first-order-logic
logic first-order-logic
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
asked Jan 18 at 18:06
LoMaPhLoMaPh
8111916
8111916
add a comment |
add a comment |
1 Answer
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$begingroup$
Your last line is right: "$varphi$ and $psi$ are $T$-equivalent" (which I think is the clearest way to phrase it in natural language) means "$T,varphimodelspsi$ and $T,psimodelsvarphi$."
(Meanwhile, "$Tmodels (varphimodelspsi wedgepsimodelsvarphi)$" doesn't actually mean anything - we can't "iterate $models$.")
However, there are a couple other (equivalent) ways to formulate this:
By the completeness theorem, that's equivalent to "$T,varphivdashpsi$ and $T,psivdashvarphi$" - that is, we can replace semantic entailment with proof.
By the deduction theorem, that's equivalent to "$Tvdashvarphileftrightarrowpsi$."
The deduction theorem also has a (much simpler to prove) semantic counterpart: it's easy to show that $T,thetamodelseta$ iff $Tmodelsthetarightarrow eta$, and it follows that ($T,thetamodelseta$ and $T,etamodelstheta$) iff $Tmodelsthetaleftrightarroweta$. So finally, yet another equivalent way to express "$varphi$ and $psi$ are $T$-equivalent" - and in my opinion the snappiest really - is $$"Tmodelsvarphileftrightarrowpsi."$$ Indeed,it's this last one that I consider the definition of "$varphi$ and $psi$ are $T$-equivalent," and I'd consider the equivalences with the other three versions to be (easily proved) theorems.
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
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1 Answer
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1 Answer
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$begingroup$
Your last line is right: "$varphi$ and $psi$ are $T$-equivalent" (which I think is the clearest way to phrase it in natural language) means "$T,varphimodelspsi$ and $T,psimodelsvarphi$."
(Meanwhile, "$Tmodels (varphimodelspsi wedgepsimodelsvarphi)$" doesn't actually mean anything - we can't "iterate $models$.")
However, there are a couple other (equivalent) ways to formulate this:
By the completeness theorem, that's equivalent to "$T,varphivdashpsi$ and $T,psivdashvarphi$" - that is, we can replace semantic entailment with proof.
By the deduction theorem, that's equivalent to "$Tvdashvarphileftrightarrowpsi$."
The deduction theorem also has a (much simpler to prove) semantic counterpart: it's easy to show that $T,thetamodelseta$ iff $Tmodelsthetarightarrow eta$, and it follows that ($T,thetamodelseta$ and $T,etamodelstheta$) iff $Tmodelsthetaleftrightarroweta$. So finally, yet another equivalent way to express "$varphi$ and $psi$ are $T$-equivalent" - and in my opinion the snappiest really - is $$"Tmodelsvarphileftrightarrowpsi."$$ Indeed,it's this last one that I consider the definition of "$varphi$ and $psi$ are $T$-equivalent," and I'd consider the equivalences with the other three versions to be (easily proved) theorems.
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
add a comment |
$begingroup$
Your last line is right: "$varphi$ and $psi$ are $T$-equivalent" (which I think is the clearest way to phrase it in natural language) means "$T,varphimodelspsi$ and $T,psimodelsvarphi$."
(Meanwhile, "$Tmodels (varphimodelspsi wedgepsimodelsvarphi)$" doesn't actually mean anything - we can't "iterate $models$.")
However, there are a couple other (equivalent) ways to formulate this:
By the completeness theorem, that's equivalent to "$T,varphivdashpsi$ and $T,psivdashvarphi$" - that is, we can replace semantic entailment with proof.
By the deduction theorem, that's equivalent to "$Tvdashvarphileftrightarrowpsi$."
The deduction theorem also has a (much simpler to prove) semantic counterpart: it's easy to show that $T,thetamodelseta$ iff $Tmodelsthetarightarrow eta$, and it follows that ($T,thetamodelseta$ and $T,etamodelstheta$) iff $Tmodelsthetaleftrightarroweta$. So finally, yet another equivalent way to express "$varphi$ and $psi$ are $T$-equivalent" - and in my opinion the snappiest really - is $$"Tmodelsvarphileftrightarrowpsi."$$ Indeed,it's this last one that I consider the definition of "$varphi$ and $psi$ are $T$-equivalent," and I'd consider the equivalences with the other three versions to be (easily proved) theorems.
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
add a comment |
$begingroup$
Your last line is right: "$varphi$ and $psi$ are $T$-equivalent" (which I think is the clearest way to phrase it in natural language) means "$T,varphimodelspsi$ and $T,psimodelsvarphi$."
(Meanwhile, "$Tmodels (varphimodelspsi wedgepsimodelsvarphi)$" doesn't actually mean anything - we can't "iterate $models$.")
However, there are a couple other (equivalent) ways to formulate this:
By the completeness theorem, that's equivalent to "$T,varphivdashpsi$ and $T,psivdashvarphi$" - that is, we can replace semantic entailment with proof.
By the deduction theorem, that's equivalent to "$Tvdashvarphileftrightarrowpsi$."
The deduction theorem also has a (much simpler to prove) semantic counterpart: it's easy to show that $T,thetamodelseta$ iff $Tmodelsthetarightarrow eta$, and it follows that ($T,thetamodelseta$ and $T,etamodelstheta$) iff $Tmodelsthetaleftrightarroweta$. So finally, yet another equivalent way to express "$varphi$ and $psi$ are $T$-equivalent" - and in my opinion the snappiest really - is $$"Tmodelsvarphileftrightarrowpsi."$$ Indeed,it's this last one that I consider the definition of "$varphi$ and $psi$ are $T$-equivalent," and I'd consider the equivalences with the other three versions to be (easily proved) theorems.
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
Your last line is right: "$varphi$ and $psi$ are $T$-equivalent" (which I think is the clearest way to phrase it in natural language) means "$T,varphimodelspsi$ and $T,psimodelsvarphi$."
(Meanwhile, "$Tmodels (varphimodelspsi wedgepsimodelsvarphi)$" doesn't actually mean anything - we can't "iterate $models$.")
However, there are a couple other (equivalent) ways to formulate this:
By the completeness theorem, that's equivalent to "$T,varphivdashpsi$ and $T,psivdashvarphi$" - that is, we can replace semantic entailment with proof.
By the deduction theorem, that's equivalent to "$Tvdashvarphileftrightarrowpsi$."
The deduction theorem also has a (much simpler to prove) semantic counterpart: it's easy to show that $T,thetamodelseta$ iff $Tmodelsthetarightarrow eta$, and it follows that ($T,thetamodelseta$ and $T,etamodelstheta$) iff $Tmodelsthetaleftrightarroweta$. So finally, yet another equivalent way to express "$varphi$ and $psi$ are $T$-equivalent" - and in my opinion the snappiest really - is $$"Tmodelsvarphileftrightarrowpsi."$$ Indeed,it's this last one that I consider the definition of "$varphi$ and $psi$ are $T$-equivalent," and I'd consider the equivalences with the other three versions to be (easily proved) theorems.
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
answered Jan 18 at 18:18
Noah SchweberNoah Schweber
124k10150287
124k10150287
add a comment |
add a comment |
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