¬p ⊬ ⎕(p → q): Where's the mistake in my proof?

Question

My professor noted on one of his slides that ¬p ⊬ ⎕(p → q). Intuitively, this seems correct; however, I can only prove that it is false. I suspect I've made a mistake in my proof. Where have I gone wrong?

1. ¬(¬p→⎕(p → q)), w0✓
2. ¬p ∧ ◊¬(p → q), w0✓
3. ¬p, w0
4. ◊¬(p → q), w0✓w1
5. w0Rw1
6. ¬(p → q),w1✓
7. p∧¬q, w1✓
8. p, w1
9. ¬q, w1
10. ⊥: (3.¬p w0, 5.w0Rw1, 8. p w1)

It is false that it is false that ¬p→⎕(p → q), so it is true that ¬p→⎕(p → q). If it is true that ¬p→⎕(p → q), then it cannot be the case that ¬p ⊬ ⎕(p → q).

Answer 1

You have $\neg p$ back in $w_0$, and $p$ in $w_1$. But there's no contradiction between those two, even if $w_0Rw_1$!

It is absolutely fundamental that a propositional variable can take different values at different worlds, even if those worlds are related by the accessibility relation.