Let $D$ be an effectively generated set of recursive ordinals the 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 $FALSE$?