I've been thinking about the unexpected hanging paradox again.

Until today, I always thought the right solution was given by Fitch in "A Goedelized Formulation of the Prediction Paradox". Let's define a sentence in Peano arithmetic called Surprise, which will refer to five other sentences Mon, Tue, Wed, Thu, Fri. Surprise will be defined recursively using the diagonal lemma, as a conjunction of these sentences:

1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.

2) If Mon is true then "Surprise implies Mon" isn't provable.

3) If Tue is true then "Surprise implies {Mon or Tue}" isn't provable.

4) If Wed is true then "Surprise implies {Mon or Tue or Wed}" isn't provable.

5) If Thu is true then "Surprise implies {Mon or Tue or Wed or Thu}" isn't provable.

6) If Fri is true then "Surprise implies {Mon or Tue or Wed or Thu or Fri}" isn't provable.

All self-references are legal because they occur inside "provable" quotes. Now it's easy to prove that Surprise is false, no matter what Mon, Tue, Wed, Thu and Fri say. So the teacher is lying and that seems to be the end of the paradox.

But it just occurred to me that there's a much simpler solution that doesn't require Gödel encoding. Let's define a sentence Surprise in naive logic with self-reference, as a conjunction of these:

1) Exactly one of Mon, Tue, Wed, Thu, Fri is true.

2) If Mon is true then "Surprise implies Mon" is false.

3) If Tue is true then "Surprise implies {Mon or Tue}" is false.

4) If Wed is true then "Surprise implies {Mon or Tue or Wed}" is false.

5) If Thu is true then "Surprise implies {Mon or Tue or Wed or Thu}" is false.

6) If Fri is true then "Surprise implies {Mon or Tue or Wed or Thu or Fri}" is false.

Now it's even easier to prove that Surprise can't be true (it's either false or indeterminate). The proof is similar to Fitch's reasoning, but without the complicated machinery. It seems to me that it resolves the paradox just as effectively, no?