Mixing
Reference request: Reflection as a single axiom interpreting the whole of ZFC?
$begingroup$
Working in first order languge $mathcal L(in, W)$, where $W$ is a constant symbol.
Reflection: if $varphi$ is a formula in $mathcal L(in)$, in which $x$ is free, and $vec{p}$ is the string of all of its parameters, and if $psi$ is a formula in which $z$ is free and $y$ not free, then all closures of: $$vec{p} in W wedge exists x (varphi) to exists x [varphiwedge exists y in W forall z (z in y leftrightarrow z in x wedge psi) ] $$; are axioms.
Now with Extensionality this would interpret all axioms of $text{ZFC}$ since it would trivially interpret Harvey Friedman $text{K(W)}$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.
Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?
reference-request set-theory first-order-logic
asked Jan 25 at 13:45
ZuhairZuhair
345212
$endgroup$
add a comment |
$begingroup$
Working in first order languge $mathcal L(in, W)$, where $W$ is a constant symbol.
Reflection: if $varphi$ is a formula in $mathcal L(in)$, in which $x$ is free, and $vec{p}$ is the string of all of its parameters, and if $psi$ is a formula in which $z$ is free and $y$ not free, then all closures of: $$vec{p} in W wedge exists x (varphi) to exists x [varphiwedge exists y in W forall z (z in y leftrightarrow z in x wedge psi) ] $$; are axioms.
Now with Extensionality this would interpret all axioms of $text{ZFC}$ since it would trivially interpret Harvey Friedman $text{K(W)}$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.
Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?
reference-request set-theory first-order-logic
asked Jan 25 at 13:45
ZuhairZuhair
345212
$endgroup$
$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10
add a comment |
$begingroup$
Working in first order languge $mathcal L(in, W)$, where $W$ is a constant symbol.
Reflection: if $varphi$ is a formula in $mathcal L(in)$, in which $x$ is free, and $vec{p}$ is the string of all of its parameters, and if $psi$ is a formula in which $z$ is free and $y$ not free, then all closures of: $$vec{p} in W wedge exists x (varphi) to exists x [varphiwedge exists y in W forall z (z in y leftrightarrow z in x wedge psi) ] $$; are axioms.
Now with Extensionality this would interpret all axioms of $text{ZFC}$ since it would trivially interpret Harvey Friedman $text{K(W)}$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.
Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?
reference-request set-theory first-order-logic
asked Jan 25 at 13:45
ZuhairZuhair
345212
$endgroup$
Working in first order languge $mathcal L(in, W)$, where $W$ is a constant symbol.
Reflection: if $varphi$ is a formula in $mathcal L(in)$, in which $x$ is free, and $vec{p}$ is the string of all of its parameters, and if $psi$ is a formula in which $z$ is free and $y$ not free, then all closures of: $$vec{p} in W wedge exists x (varphi) to exists x [varphiwedge exists y in W forall z (z in y leftrightarrow z in x wedge psi) ] $$; are axioms.
Now with Extensionality this would interpret all axioms of $text{ZFC}$ since it would trivially interpret Harvey Friedman $text{K(W)}$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.
Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?
reference-request set-theory first-order-logic
reference-request set-theory first-order-logic
asked Jan 25 at 13:45
ZuhairZuhair
345212
asked Jan 25 at 13:45
ZuhairZuhair
345212
edited Jan 27 at 19:41
Zuhair
asked Jan 25 at 13:45
ZuhairZuhair
345212
asked Jan 25 at 13:45
ZuhairZuhair
345212
asked Jan 25 at 13:45
ZuhairZuhair
345212
345212
$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10
add a comment |
$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10
$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10
add a comment |
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$begingroup$
Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes."
$endgroup$
– Noah Schweber
Jan 25 at 18:51
$begingroup$
To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others
$endgroup$
– Zuhair
Jan 25 at 19:10