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Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0=1$, whis is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem, because this is a theory that is $k$-inconsistent but does not $k$-prove its own $k$-consistency, contrary to the proposed equivalence in your question.

Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0=1$, which is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem, because this is a theory that is $k$-inconsistent but does not $k$-prove its own $k$-consistency, contrary to the proposed equivalence in your question.

Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0\neq 1$, whis is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem.

Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0=1$, whis is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem, because this is a theory that is $k$-inconsistent but does not $k$-prove its own $k$-consistency, contrary to the proposed equivalence in your question.

Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But unfortunately, this is not a $k$-proof of its own consistency, so it doesn't quite answer your second question... In fact, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

Nice questions!

For the first question, I claim that every theory $T$ is $k$-incomplete in your sense for every finite $k$. This is because every statement appears explicitly as a part of any proof of it, and when $\phi$ is any very long theorem of $T$ in comparison with $k$, then any proof of $\phi$ has at least as many symbols in it as $\phi$ does, and so by your measures $\phi$ has no proof of size $k$ and hence is $k$-Goedelian by your definition. Thus, for any given $k$, the theory $T$ is $k$-incomplete. In particular, every theory becomes $k$-incomplete.

For the second question, it seems that any decent consistent theory will prove its own $k$-consistency for any particular $k$. For example, if $T$ is consistent and extends $PA$, and probably even extending $Q$ suffices, then since $T$ really does not have any proofs of contradictions, it has none of size $k$, and since there are only finitely many proofs of this size, which can be enumerated and known to be an exhaustive list within the theory $T$, it follows that $T$ will prove that there is no proof of a contradiction in $T$ of size at most $k$. That is, for each $k$ separately, $T$ proves its own $k$-consistency.

But since this is not a $k$-proof of its own consistency, it doesn't quite answer your second question. But it shows that you have to pay attention to the precise measure of $k$-provability. For example, it isn't even clear that a $k$-inconsistent theory will necessarily $k$-prove much, since perhaps the $k$-proof of a contradiction already uses up most of the $k$ symbols, leaving little room for further deductions from that contradiction.

In light of this, we can make a negative answer to your second question as follows. Let $T$ be the theory PA + $0\neq 1$, whis is $k$-inconsistent for a very small value of $k$, perhaps $25$ or so, since the extra axiom directly contradicts an axiom of PA. But this $k$ is much too small to even state the $k$-consistency of $T$, and so $T$ will not $k$-prove its own $k$-consistency. So it violates the second $k$-incompleteness theorem.