Usually, having more comprehension axiom means the more you can prove. We wonder if the converse can be the case.


Is there a natural problem $\mathsf{P}$ so that $\mathsf{P}+\neg(\Gamma-\mathsf{ComprehensionAxiom})$ implies
$\mathsf{Q}$, but $\mathsf{P}$ does not implies $\mathsf{Q}$.


Also I'd like to exclude some trivial case such as $\mathsf{Q}$ is $\neg(\Gamma-\mathsf{ComprehensionAxiom})$. We can accept $\mathsf{Q}$ to be some Comprehension Axiom, but not the negation of them.
The implication is not necessary over $\mathsf{RCA}$ and $\Gamma$ can be any set of formulas.



Comprehension axiom that helps in the opposite direction
David Belanger's work is relevant.


The principle $\mathsf{WKL\_0}$ (= "Every infinite subtree of $2^{<\omega}$ has an infinite path") is not literally a comprehension principle, so its negation can be used as a choice of $\mathsf{Q}$. Belanger [showed](https://www.jstor.org/stable/43303769) that over $\mathsf{RCA\_0}$, the principle $(\star)\equiv$ "There is a complete theory with finitely many models and a single nonprincipal type" is equivalent to $\mathsf{ACA\_0\vee\neg WKL\_0}$. Consequently, if we take as our base theory $\mathsf{RCA\_0+(\star)}$, we get $\neg\mathsf{ACA\_0}\implies\mathsf{Q}$.


What if we *do* want to consider $\mathsf{WKL\_0}$ as a kind of comprehension principle (it's one of the "Big Five" after all), so that we can't set $\mathsf{Q}=\neg\mathsf{WKL\_0}$? Well, in this case another result of Belanger is relevant: that a rather complicated model-theoretic principle (see Theorem 2.23 [here](https://www.sciencedirect.com/science/article/pii/S0168007215000330?via%3Dihub)), which I'll refer to as $(\dagger)$ here, implies over $\mathsf{RCA\_0}$ the disjunction $\mathsf{WKL\_0\vee I\Sigma\_2}$. Consequently, over $\mathsf{RCA\_0}$ the negation of $\mathsf{WKL\_0}$ (which we're now considering a comprehension principle!), plus the principle $(\dagger)$, implies $\mathsf{I\Sigma\_2}$ - which is not itself a consequence of $\mathsf{RCA\_0}+(\dagger)$ so this is nontrivial.


