Why is a relation that is $Sigma-$representable not an IFF condition?
I was reading these notes and the definition of $Sigma-$representability of a relation. It's as followings. $R$ is $Sigma$ representable if $forall a in mathbf N^m$ (where $a_i = S^{a_i} 0$ and $S$ is the successor function):
- $R(a) implies Sigma vdash varphi(a)$
- $neg R(a) implies Sigma vdash neg varphi(a)$
however for some reason the definition does not apply both ways automatically (which is what I would have expected). It does when $Sigma$ is consistent (for some reason).
- $R(a) iff Sigma vdash varphi(a)$
- $neg R(a) iff Sigma vdash neg varphi(a)$
Why does that only happen when $Sigma$ is consistent (and thus can be completed easily by Lindenbaum).
logic first-order-logic predicate-logic
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
add a comment |
I was reading these notes and the definition of $Sigma-$representability of a relation. It's as followings. $R$ is $Sigma$ representable if $forall a in mathbf N^m$ (where $a_i = S^{a_i} 0$ and $S$ is the successor function):
- $R(a) implies Sigma vdash varphi(a)$
- $neg R(a) implies Sigma vdash neg varphi(a)$
however for some reason the definition does not apply both ways automatically (which is what I would have expected). It does when $Sigma$ is consistent (for some reason).
- $R(a) iff Sigma vdash varphi(a)$
- $neg R(a) iff Sigma vdash neg varphi(a)$
Why does that only happen when $Sigma$ is consistent (and thus can be completed easily by Lindenbaum).
logic first-order-logic predicate-logic
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
add a comment |
I was reading these notes and the definition of $Sigma-$representability of a relation. It's as followings. $R$ is $Sigma$ representable if $forall a in mathbf N^m$ (where $a_i = S^{a_i} 0$ and $S$ is the successor function):
- $R(a) implies Sigma vdash varphi(a)$
- $neg R(a) implies Sigma vdash neg varphi(a)$
however for some reason the definition does not apply both ways automatically (which is what I would have expected). It does when $Sigma$ is consistent (for some reason).
- $R(a) iff Sigma vdash varphi(a)$
- $neg R(a) iff Sigma vdash neg varphi(a)$
Why does that only happen when $Sigma$ is consistent (and thus can be completed easily by Lindenbaum).
logic first-order-logic predicate-logic
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
I was reading these notes and the definition of $Sigma-$representability of a relation. It's as followings. $R$ is $Sigma$ representable if $forall a in mathbf N^m$ (where $a_i = S^{a_i} 0$ and $S$ is the successor function):
- $R(a) implies Sigma vdash varphi(a)$
- $neg R(a) implies Sigma vdash neg varphi(a)$
however for some reason the definition does not apply both ways automatically (which is what I would have expected). It does when $Sigma$ is consistent (for some reason).
- $R(a) iff Sigma vdash varphi(a)$
- $neg R(a) iff Sigma vdash neg varphi(a)$
Why does that only happen when $Sigma$ is consistent (and thus can be completed easily by Lindenbaum).
logic first-order-logic predicate-logic
logic first-order-logic predicate-logic
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
asked Nov 30 '18 at 2:28
PinocchioPinocchio
1,88021854
1,88021854
add a comment |
add a comment |
1 Answer
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If $Sigma$ is inconsistent, it proves everything, so $Sigmavdash varphi(mathbf a)$ and $Sigma vdash lnot varphi(mathbf a)$ always hold, whereas at most one of $R(a)$ or $lnot R(a)$ hold, so it is impossible for both $leftarrow$ directions to hold for any $a,$ let alone all of them.
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
add a comment |
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1 Answer
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1 Answer
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If $Sigma$ is inconsistent, it proves everything, so $Sigmavdash varphi(mathbf a)$ and $Sigma vdash lnot varphi(mathbf a)$ always hold, whereas at most one of $R(a)$ or $lnot R(a)$ hold, so it is impossible for both $leftarrow$ directions to hold for any $a,$ let alone all of them.
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
add a comment |
If $Sigma$ is inconsistent, it proves everything, so $Sigmavdash varphi(mathbf a)$ and $Sigma vdash lnot varphi(mathbf a)$ always hold, whereas at most one of $R(a)$ or $lnot R(a)$ hold, so it is impossible for both $leftarrow$ directions to hold for any $a,$ let alone all of them.
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
add a comment |
If $Sigma$ is inconsistent, it proves everything, so $Sigmavdash varphi(mathbf a)$ and $Sigma vdash lnot varphi(mathbf a)$ always hold, whereas at most one of $R(a)$ or $lnot R(a)$ hold, so it is impossible for both $leftarrow$ directions to hold for any $a,$ let alone all of them.
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
If $Sigma$ is inconsistent, it proves everything, so $Sigmavdash varphi(mathbf a)$ and $Sigma vdash lnot varphi(mathbf a)$ always hold, whereas at most one of $R(a)$ or $lnot R(a)$ hold, so it is impossible for both $leftarrow$ directions to hold for any $a,$ let alone all of them.
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
edited Nov 30 '18 at 2:43
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Nov 30 '18 at 2:42
spaceisdarkgreenspaceisdarkgreen
32.5k21753
32.5k21753
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
add a comment |
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
I see I missed that if its not consistent it must be inconsistent...law of excluded middle in the meta logic?
– Pinocchio
Nov 30 '18 at 2:43
add a comment |
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