There’s a flaw in your proof—the step from “The machine will necessarily stop” to “This solves the halting problem” is unjustified.
Despite the flaw, I agree with your general conclusion—there are and will be challenging as well as unsolvable problems.
It’s justified. If a machine halts, then there’s a proof of that in PA (simply the list of steps it performs before halting). Therefore, “independent of PA” implies “doesn’t halt”. Therefore, a provability decider is the same as a halting decider.
With the added steps, yes, there is a proof.
There’s a flaw in your proof—the step from “The machine will necessarily stop” to “This solves the halting problem” is unjustified.
Despite the flaw, I agree with your general conclusion—there are and will be challenging as well as unsolvable problems.
It’s justified. If a machine halts, then there’s a proof of that in PA (simply the list of steps it performs before halting). Therefore, “independent of PA” implies “doesn’t halt”. Therefore, a provability decider is the same as a halting decider.
With the added steps, yes, there is a proof.