Is the 1-consistency of $PA$ necessary to prove that $\diamond^{m} \top\implies \diamond^{n} \top$ is false if $m<n$?

Question

In The logic of provability, by G. Boolos, there is a remark in chapter 7 saying that $\diamond^{m} \top\implies \diamond^{n} \top$ is false if $m<n$ (unless $PA$ is 1-inconsistent).

Now, it seems to me that the parenthetical expression is not necessary, since earlier in the chapter we saw a demonstration of the arithmetical completeness theorem of $GL$ for constant sentences (sentences without sentence letters), which did require simple consistency of $PA$ to work, instead of 1-consistency.

Since the expression in question is letterless, and it is easy to see that it is not a theorem of $GL$, attending to its trace, then we should be able to conclude that $PA$ doesn't prove that either appealing to mere consistency.

Am I right? Is there something I am misunderstanding?

Answer 1

Nevermind, I just realized that in the proof of arithmetical completeness the author also needs to assume 1-consistency (To show that Bew(False) is not provable).

Answer 2

Hmm your reasoning seems weird. Take any theory $T$ that satisfies the provability conditions. Take any $m,n \in \mathbb{N}$ such that $m<n$. Then by Lob's theorem, if $T \vdash \square^n \bot \to \square^{n-1} \bot$ then $T \vdash \square^{n-1} \bot$. Note that $T \vdash \square^m \bot \to \square^{n-1} \bot$ by using (D3) for $m>0$. Thus if $T \vdash \square^n \bot \to \square^m \bot$ we would have $T \vdash \square^{n-1} \bot$. Now if $n = 1$ then $T$ is actually inconsistent. But if $n > 1$ then $T$ must be $Σ_1$-unsound since otherwise $T \vdash \bot$ by iteratively "removing the $\square$".

Since there are these two cases, I'm not sure how you could conclude that one always needs to assume $Σ_1$-soundness ($1$-consistency).