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Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.

Note: This post modified the earlier approach before the [HOLD] which was erroneous in using indefinable ordinal indices, here this is fixed to index the theories after an ordinal notation that is effectively recursively generated.

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

   Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.

Note: This post modified the earlier approach before the [HOLD] which was erroneous in using indefinable ordinal indices, here this is fixed to index the theories after an ordinal notation that is effectively recursively generated.

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.

Note: This post modified the earlier approach before the [HOLD] which was erroneous in using indefinable ordinal indices, here this is fixed to index the theories after an ordinal notation that is effectively recursively generated.

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

• $T_0 = T$,
• $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
• $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.