TITLE: Is there a statement independent from PA and does not increase the consistency strength?
QUESTION [6 upvotes]: There are many known independence results of $\mathsf{PA}$, for example, Goodstein theorem, Paris-Harrington theorem, and the reflection principle. But these examples imply the consistency of $\mathsf{PA}$.
I guess not all independent statements of $\mathsf{PA}$ imply the consistency of $\mathsf{PA}$. I have tried to find such examples, but I can't even prove the existence of such a proposition. I would appreciate your help.

REPLY [3 votes]: The Rosser sentence $\rho$ is an example. 
The Rosser sentence $\rho$ for a theory T (containing some weak arithmetic) is the assertion that for every proof of $\rho$ in $T$, there is a shorter proof of $\neg\rho$, that is, a proof with a smaller Gödel code. (One can use the fixed-point lemma to handle the apparent self-reference.)
If $T$ is consistent, then $\rho$ is independent of $T$, by the following well-known argument. If $T\vdash\rho$, then $T$ would not also prove $\neg\rho$, but it would have to, in light of what $\rho$ expresses. So $T\not\vdash\rho$. Similarly, if $T\vdash\neg\rho$, then $T$ would have to prove that there is a shorter proof of $\rho$, which it cannot because it is consistent. 
Thus, we've proved $\text{Con}(T)\to\text{Con}(T+\rho)\wedge\text{Con}(T+\neg\rho)$, in a very weak theory. Thus, $T+\rho$ and $T+\neg\rho$ are equiconsistent with $T$. So $\rho$ is a sentence that is indepdnent of $T$, without increasing consistency strength. 
In particular, the Rosser sentence $\rho$ is strictly weaker than $\text{Con}(T)$.