show/hide this revision's text 5 redacted old answer

Note: The top part answers an old version of the question, which is now irrelevant.

Given a axiomatizable theory T of arithmetic, the set of all statements independent of T is the complement of a computably enumerable set. When nonempty (e.g. when T extends PA) this set is countably infinite. (Trivially, if φ is such then so is φ∧∃x(x=x), etc.) However, there is no general algorithm to produce the shortest element of such a set, never mind counting them. (The requirement that the sentence be true is negligible since negation only adds a constant number of symbols depending on syntactic conventions.)

That said, some variants of your question have been actively studied. Hilbert's Tenth Problem says that there are Diophantine equations that have no integer solutions, but this fact is not provable from T. The question of the minimum number variables and minimum degree such diophantine equations have been studied. Over Z, Skolem showed that degree 4 is sufficient and Zhi-Wei Sun showed that 11 variables is sufficient. It is unknown whether these results are optimal.

Now that I reread your question, I think you wanted to have infinitely many logically inequivalent statements each of which is independent of T. This is true when T has no axiomatizable complete extension, which is guaranteed Gödel's Theorem when T is a consistent axiomatizable theory that extends PA.

Indeed, if there were only finitely many statements φ1,...,φk independent of T, up to T-provable equivalence. Then we could get an axiomatizable extension of T by adding to T each such φi or its negation ¬φi while maintaining consistency. (For example, when the standard model satisfies T, we can pick whichever is true in the standard model.) Since we're only adding finitely many new axioms, the result would be a complete an axiomatizable complete theory even if our finitely many decisions were very complex; this would contradict Gödel's Theorem.
show/hide this revision's text 4 typo

Given a axiomatizable theory T of arithmetic, the set of all statements independent of T is the complement of a computably enumerable set. When nonempty (e.g. when T extends PA) this set is countably infinite. (Trivially, if φ is such then so is φ∧∃x(x=x), etc.) However, there is no general algorithm to produce the shortest element of such a set, never mind counting them. (The requirement that the sentence be true is negligible since negation only adds a constant number of symbols depending on syntactic conventions.)

That said, some variants of your question have been actively studied. Hilbert's Tenth Problem says that there are Diophantine equations that have no integer solutions, but this fact is not provable from T. The question of the minimum number variables and minimum degree such diophantine equations have been studied. Over Z, Skolem showed that degree 4 is sufficient and Zhi-Wei Sun showed that 11 variables is sufficient. It is unknown whether these results are optimal.

Now that I reread your question, I think you wanted to have infinitely many logically inequivalent statements each of which is independent of T. This is true when T has no axiomatizable complete extension, which is guaranteed Gödel's Theorem when T is a consistent axiomatizable theory that extends PA.

Indeed, if there were only finitely many statements φ1,...,φk statements independent of T, up to T-provable equivalence. Then we could get an axiomatizable extension of T by adding to T each such φi or its negation ¬φi while maintaining consistency. (For example, when the standard model satisfies T, we can pick whichever is true in the standard model.) Since we're only adding finitely many new axioms, the result would be a complete axiomatizable theory even if our finitely many decisions were very complex; this would contradict Gödel's Theorem.
show/hide this revision's text 3 reworded for clarity

Given a axiomatizable theory T of arithmetic, the set of all statements independent of T is the complement of a computably enumerable set. When nonempty (e.g. when T extends PA) this set is countably infinite. (Trivially, if φ is such then so is φ∧∃x(x=x), etc.) However, there is no general algorithm to produce the shortest element of such a set, never mind counting them. (The requirement that the sentence be true is negligible since negation only adds a constant number of symbols depending on syntactic conventions.)

That said, some variants of your question have been actively studied. Hilbert's Tenth Problem says that there are Diophantine equations that have no integer solutions, but this fact is not provable from T. The question of the minimum number variables and minimum degree such diophantine equations have been studied. Over Z, Skolem showed that degree 4 is sufficient and Zhi-Wei Sun showed that 11 variables is sufficient. It is unknown whether these results are optimal.

Now that I reread your question, I think you wanted to have infinitely many logically inequivalent statements each of which is independent of T. This is true when T has no axiomatizable complete extension. (This , which is guaranteed Gödel's Theorem when T is a consistent axiomatizable theory that extends PAby Gödel's Theorem.) .

Indeed, if there were only finitely many statements φ1,...,φk statements independent of T up to logical T-provable equivalence, then . Then we could get an axiomatizable extension of T by adding to T each such statement φi or its negation ¬φi while maintaining consistency. (For example, when the standard model satisfies T, we can pick whichever is true in the standard model.) Since we're only adding finitely many new axioms, the result would be an axiomatizable and a complete axiomatizable theory , even if our finitely many decisions were very complex; this would contradict Gödel's Theorem.
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