TITLE: Infinite descending consistency chains
QUESTION [12 upvotes]: What are some examples of consistent theories $T_i$ (extending elementary arithmetic EA) such that for $∀i∈ℕ \,\, T_i ⊢ \mathrm{Con}(T_{i+1})$?
Such theories exist; see for example An infinitely descending sequence of sound theories each proving the next consistent. However, in the link, the theories are given using (non-well-founded) self-reference, which in an intuitive informal sense, leaves one wondering what exactly is each theory asserting.
Thus, I am especially interested in natural or intuitive examples that do not use self-reference.  Here is what I have.
If we only require $\mathrm{PA} ⊬ (\mathrm{Con}(T_{i+1}) ⇒ \mathrm{Con}(T_i))$, then for an example, fix a proof system that is constructively provably in $\mathrm{EA}$ polynomial time equivalent to sequent calculus.  Let $n$ (provably in EA) be the number of bits in the shortest inconsistency in $\mathrm{PA}$ if there is any.  If $T_i = \mathrm{EA} + \mathrm{Con}(\mathrm{PA}) ∨ 2^i ∤ ⌊\log \log \log n⌋$, then for all $i$, $T_i ⊢ T_{i+1}$ and $\mathrm{PA} ⊬ (\mathrm{Con}(T_{i+1}) ⇒ \mathrm{Con}(T_i))$.  (To get the unpredictability of $n$, the proof relies on $\mathrm{PA}$ provability of $\mathrm{PA}$ bounded consistency in polynomial time, and the impossibility of a subpolynomial proof length of bounded consistency.)
To get $T_i ⊢ \mathrm{Con}(T_{i+1})$, I have two potential natural examples, but I do not have a proof that they work.
Update: Per Fedor Pakhomov's answer, the original constructions (given in brackets for reference) did not work; below are the modified versions.  See the answers for constructions that have been shown to work; the answers complement each other.
Construction 1:  Let $S_0 = \mathrm{PA}$ and $S_{i+1} = \mathrm{PA} + \mathrm{Con}(S_i)$, and let $h_{ε_0}$ be the $ε_0$ function in the Hardy hierarchy.
Set $T_i = \mathrm{PA} + ∀n \,\, (h_{ε_0 2}(i+n) \text{ exists} ⇒ \mathrm{Con}(S_n))$ [the original used $h_{ε_0}$, but perhaps even $h_{ε_0 2} = λx.h_{ε_0}(h_{ε_0}(x))$ is too small]
Construction 2: Let $S_0 = \mathrm{PA}$ and $S_{i+1} = \mathrm{PA} + \mathrm{Con}(S_i)$.  We can arrange that $⌈\mathrm{Con}(S_i)⌉$ is $i^{O(1)}$, but we only need $2^{2^{2^{O(i)}}}$.  Let $\mathrm{Con}_m(A)$ mean that $A$ has no inconsistency shorter than $m$ bits.
Set $T_i = \mathrm{PA} + ∀n \, (\mathrm{Con}_{2^{n^i}}(S_{n+i}) ⇒ \mathrm{Con}(S_n))$ [the original used $S_n$ in place of $S_{n+i}$].
Note: If we can show in $\mathrm{PA}$ that each $T_i$ (for $i>0$) is consistent relative to $S=∪S_i$, then the construction works because if $S$ is inconsistent, then working in $T_i$, $T_{i+1}$ is equivalent to $S_{i'}$ for a low enough $i'$ so as to be consistent.
Also, the well-ordering of $Π^0_1$ ordinals imposes fundamental limits on descending consistency chains. I suspect that for sound $T$, $∃i \, T_i ⊬ ∀j \, (T_j ⊢ \mathrm{Con}(T_{j+1}))$.

REPLY [3 votes]: Informed by Fedor Pakhomov's excellent answer, and using "slow" iterated $Σ_1$-soundness, here is an infinite sequence of sound theories $T$ such that $T_i$ = PA + 1-Con($T_{i+1}$).
Set $T_i$ = PA + "$h'_{g(n)∸i}(n)$ is total"
using fast-growing or Hardy hierarchy $h$ (either one works), with $h'_α$ being $h_{ε_α}$, and $g(n) = {h'_{ω+1}}^{-1}(n)$, where $f^{-1}(n) = \max(m: ∀m'<m \,\, f(m')<n)$ (or any reasonable variation on this).   Also, $a∸b = \max(a-b,0)$.
Alternatively, almost any reasonable monotonic total recursive $g$ that is sufficiently slow-growing but tending to infinity works, but note that $T$ depends on $g$. The construction also generalizes to other theories extending $Σ^0_1$-PA (the proof uses $Σ^0_1$ induction) by replacing $h'_m(n)$ with a function corresponding to $1+m$ iterations of $Σ_1$-soundness.
Provably in PA, if $g$ is unbounded (equivalently, $h'_{ω+1}$ is total), then $T_i$ (and even PA + "$h'_ω$ is total") is $Σ_1$-sound.  Thus, it suffices to prove that for each $i$, PA + "$g$ is bounded" proves "$h'_{g(n)∸i}(n)$ is total" $⇔$ 1-Con($T_{i+1}$). (However, the quantification over $i$ will be unprovable even in $T_0$).
Working in PA + "$g$ is bounded", let $k$ be the the maximum value of $g$. Since $k$ is (externally) nonstandard, $k>i$.  Because 1-Con is unaffected by true $Π^0_1$ statements, in 1-Con(...), we can freely assume/assert that $k$ is the maximum value of $g$.  Thus, 1-Con($T_{i+1}$) $⇔$ 1-Con(PA + "$h'_{k-(i+1)}(n)$ is total") $⇔$ "$h'_{k-i}(n)$ is total"  (the latter follows from standard results in ordinal analysis), as required.
Addendum ($Σ^1_1$-soundness):  Per the linked paper in Fedor Pakhomov's answer, the $Σ^0_1$ soundness cannot be improved to $Σ^0_2$ (equivalently $Π^0_3$) for c.e. (in $i$) $T$.  However, if $T$ is not required to be computable enumerable in $i$, there is an infinite sequence $T_i$ of sound c.e. extensions of Z$_2$ such that each $T_i$ proves $Σ^1_1$-soundness of $T_{i+1}$.  $T_i$ will be satisfied by every transitive model of Z$_2$ (and analogously for theories other than Z$_2$ extending $Π^1_1$-CA$_0$).
   To see this, given a recursive ordering $X$, let $T_a$ state Z$_2$ plus existence of a real $M$ such that for all $b ≤_X a$, $M_b$ is a (countably coded) $ω$ model of Z$_2$, and for all $c <_X b$, $M_c$ is an element of (the model coded by) $M_b$.  Such $M$ exists for all $a$ in the well-founded part of $X$, so if $X$ is a recursive pseudowellordering, then such $M$ must also exist for some $a'$ outside of the well-founded part of $X$.  If $a_i$ with $a_0=a'$ is an infinite $X$-descending sequence, then $T_{a_i}$ proves $Σ^1_1$-soundness $T_{a_{i+1}}$.  Also, I suspect that there are sound finitely axiomatizable $T_i = Π^1_1\text{-CA}_0 + \mathrm{RFN}_{Σ^1_1}(T_{i+1})$, and similarly with many $Σ^1_1$ strengthenings of $\mathrm{RFN}_{Σ^1_1}$.