Is there an ordering of logical systems defined by reductions? The 2019 Stack Overflow Developer Survey Results Are InStyles of logical systemsRecursively defined systems are always consistent?On Counted LanguagesLambda Calculus using $beta$-reductionsAre there formal systems that are not logical systems?Lambda calculus: Simplifying booleans with beta reductionsFormal logic versus computation?'The Computer and the Brain' - The mathematical language of the brainAre there “structure-specific” logical axiomatic systems? Do these have extra power?Ordering between formal theories by provability of consistency
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Is there an ordering of logical systems defined by reductions?
The 2019 Stack Overflow Developer Survey Results Are InStyles of logical systemsRecursively defined systems are always consistent?On Counted LanguagesLambda Calculus using $beta$-reductionsAre there formal systems that are not logical systems?Lambda calculus: Simplifying booleans with beta reductionsFormal logic versus computation?'The Computer and the Brain' - The mathematical language of the brainAre there “structure-specific” logical axiomatic systems? Do these have extra power?Ordering between formal theories by provability of consistency
$begingroup$
I am aware of the lambda cube which gives an ordering to several variants of the lambda calculus. My intuition says that this ordering should have the following property:
For logics $A,Binlambdatextrm-cube$, $Ale B$ iff there is a mapping $f$ such that formula $x$ can be inferred in $A$ iff formula $f(x)$ can be inferred in $B$. Specifically, $f(x)=x$ for languages in the $lambdatextrm-cube$, but could be non-trivial in an extended ordering.
Is this correct? And is there an extended version of this ordering that applies to a larger set of logical systems?
Then, if such an ordering exists, does it have a greatest element? And what logical system could we use to prove all of this, as (I think) we are operating at a metalogical level?
logic proof-theory lambda-calculus meta-math
asked Mar 23 at 1:39
HelloHello
63
$endgroup$
add a comment |
$begingroup$
I am aware of the lambda cube which gives an ordering to several variants of the lambda calculus. My intuition says that this ordering should have the following property:
For logics $A,Binlambdatextrm-cube$, $Ale B$ iff there is a mapping $f$ such that formula $x$ can be inferred in $A$ iff formula $f(x)$ can be inferred in $B$. Specifically, $f(x)=x$ for languages in the $lambdatextrm-cube$, but could be non-trivial in an extended ordering.
Is this correct? And is there an extended version of this ordering that applies to a larger set of logical systems?
Then, if such an ordering exists, does it have a greatest element? And what logical system could we use to prove all of this, as (I think) we are operating at a metalogical level?
logic proof-theory lambda-calculus meta-math
asked Mar 23 at 1:39
HelloHello
63
$endgroup$
$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06
add a comment |
$begingroup$
I am aware of the lambda cube which gives an ordering to several variants of the lambda calculus. My intuition says that this ordering should have the following property:
For logics $A,Binlambdatextrm-cube$, $Ale B$ iff there is a mapping $f$ such that formula $x$ can be inferred in $A$ iff formula $f(x)$ can be inferred in $B$. Specifically, $f(x)=x$ for languages in the $lambdatextrm-cube$, but could be non-trivial in an extended ordering.
Is this correct? And is there an extended version of this ordering that applies to a larger set of logical systems?
Then, if such an ordering exists, does it have a greatest element? And what logical system could we use to prove all of this, as (I think) we are operating at a metalogical level?
logic proof-theory lambda-calculus meta-math
asked Mar 23 at 1:39
HelloHello
63
$endgroup$
I am aware of the lambda cube which gives an ordering to several variants of the lambda calculus. My intuition says that this ordering should have the following property:
For logics $A,Binlambdatextrm-cube$, $Ale B$ iff there is a mapping $f$ such that formula $x$ can be inferred in $A$ iff formula $f(x)$ can be inferred in $B$. Specifically, $f(x)=x$ for languages in the $lambdatextrm-cube$, but could be non-trivial in an extended ordering.
Is this correct? And is there an extended version of this ordering that applies to a larger set of logical systems?
Then, if such an ordering exists, does it have a greatest element? And what logical system could we use to prove all of this, as (I think) we are operating at a metalogical level?
logic proof-theory lambda-calculus meta-math
logic proof-theory lambda-calculus meta-math
asked Mar 23 at 1:39
HelloHello
63
asked Mar 23 at 1:39
HelloHello
63
asked Mar 23 at 1:39
HelloHello
63
asked Mar 23 at 1:39
HelloHello
63
asked Mar 23 at 1:39
HelloHello
63
63
$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06
add a comment |
$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06
$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06
$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06
add a comment |
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$begingroup$
This mostly doesn't make sense. There are no "formulas" in the typed lambda calculi of the lambda cube. It's also unclear what you mean by "inferred". I assume something like "proven". Perhaps, via a CH lens, you want to talk about types and a mapping between types and whether they are inhabited or not. Still, you'd need some constraints on $f$ or else it's basically meaningless. What stops me from choosing $f$ as mapping every thing to just two types, one inhabited the other not, as appropriate?
$endgroup$
– Derek Elkins
Apr 1 at 9:06