The punchline at the end is worth calling out: Turing Equivalence applies at first-order computations like `Nat -> Nat`, but fails at higher-order computation l

The punchline at the end is worth calling out: Turing Equivalence applies at first-order computations like `Nat -> Nat`, but fails at higher-order computation like `(Nat -> Nat) -> {0, 1}`.