Wikipedia states that The Paris-Harrington theorem is true, but unprovable, in Peano arithmetic. The article goes on to intimate that PH can be proved in second-order arithmetic (the argument seems to be that second-order arithmetic can be used to prove PH from the infinite Ramsey theorem. A proof is given for the latter, which I assume uses Peano Arithmetic, am I right?) At any rate, I would think that if PH were true but unprovable in PA, then either PH or ¬PH would be consistent with PA. This would mean that we have two flavors of arithmetic: PA + PH, which is consistent with second-order arithmetic, and PA + ¬PH, which is not. So when the text says that PH is true but unprovable in PA, Im hearing true in second-order arithmetic but unprovable in PA, with that tacit understanding that arithmetic subsumes second-order arithmetic? Does this simply reveal a bias toward second-order arithmetic over other flavors of arithmetic such as PA + ¬PH? Or is my argument above wrong? Is it somehow possible that PA + ¬PH is contradictory, yet PH is unprovable in PA?