Let $f : \omega \to \{\text{wffs of set theory}\}$ be a function. Let $\leq_f$ be a total order on $\omega$.


Definition: $\langle f,\leq_f \rangle$ is a computable quasi-completion of ZFC if and only if

1. $f$ and $\leq_f$ are both computable
   and
   
2. for all $m,n$ in $\omega$, if $m \leq_f \; n$, then all theorems of $ZFC+f(m)$ are theorems of $ZFC+f(n)$
   and
   
3. for all sentences $s$ in the language of set theory, there exists a member $n$ of $\omega$ such that $ZFC+f(n)$ is consistent and one of $\{s,\lnot s\}$ is a theorem of $ZFC+f(n)$
   
   
   

Is it known that there is no computable quasi-completion of ZFC?

Basically, has the goal of [Ultimate L](http://mathoverflow.net/questions/46907/completion-of-zfc) been shown to be impossible?

Let $f : \omega \to \{\text{wffs of set theory}\}$ be a function. Let $\leq_f$ be a total order on $\omega$.


Definition: $\langle f,\leq_f \rangle$ is a computable quasi-completion of ZFC if and only if

1. $f$ and $\leq_f$ are both computable
   and
   
2. for all $m,n$ in $\omega$, if $m \leq_f \; n$, then all theorems of $ZFC+f(m)$ are theorems of $ZFC+f(n)$
   and
   
3. for all sentences $s$ in the language of set theory, there exists a member $n$ of $\omega$ such that $ZFC+f(n)$ is consistent and one of $\{s,\lnot s\}$ is a theorem of $ZFC+f(n)$
   
   
   

Is it known that there is no computable quasi-completion of ZFC?

Basically, has the goal of [Ultimate L](https://mathoverflow.net/questions/46907/completion-of-zfc) been shown to be impossible?