Can one use the Hilbert-Ackermann Consistency Theorem to prove the consistency of $PRA$?

Question

In his textbook Mathematical Logic, Shoenfield states the Hilbert-Ackermann Consistency Theorem as follows:

"Consistency Theorem (Hilbert-Ackermann): An open theory $T$ is inconsistent iff there is a quasi-tautology [i.e. a tautological consequence of instances of identity axioms and equality axioms--this from the first sentence of the paragraph (my comment)] which is a disjunction of negations of instances of nonlogical axioms of $T$."

He defines the term "open theory" as follows:

"[Pg. 48] A theory is open if all of its nonlogical axioms are open."

"[Pg. 36] A formula is open if it contains no quantifiers."

Since the usual formulation of $PRA$ (Primitive Recursive Arithmetic) is quantifier-free, it is, from a naive point of view, an 'open theory'.

Question: Does the Hilbert-Ackermann Consistency theorem hold for $PRA$? If so, can one (from the literature or otherwise) produce a theorem of the consistency of $PRA$ using the Hilbert-Ackermann Consistency Theorem? If not, could someone please explain to me why the theorem does not apply to $PRA$?

Answer 1

The theorem holds for PRA.

However, in order to use it to prove that PRA is consistent, you would first need to know that none of the infinitely many disjunctions of negations of instances of PRA's axioms are quasi-tautologies. At least at first glance, that doesn't look particularly easy.

Answer 2

Iirc, Shoenfield proves the theorem in PA. It is, however, a theorem of PRA. If one could use it to prove the consistency of PRA in PRA, that would prove the consistency of PRA in PRA, which is impossible by Gődel's second incompleteness theorem. One can prove the consistency of PRA in, e.g., PA quite independently of the Hilbet–Ackermann consistency theorem: if PA is consistent, so is PRA; if PA is inconsistent, then it can prove anything, hence that PRA is consistent.