Can we rule out candidate single-axiom bases for HI via Curry-Howard?

Question

The motivation for this question begins with an answer elsewhere which references Dolph Ulrich's list of single-axiom bases and unresolved candidate single-axiom bases for implicational intuitionistic logic. My interest is more the combinator side of the Curry-Howard isomorphism, so I began translating the axioms given into lambda terms.

All of the axioms that are known to work can be translated trivially. For example, HI-8 is $(p \to q) \to (((r \to (s \to p)) \to (q \to t)) \to (p \to t))$ and corresponds (after currying) to type $$(p \to q) \to ((r \to s \to p) \to q \to t) \to p \to t$$ which obviously translates into $$\lambda xyz.y(\lambda uv.z)(xz)$$

However, many of the candidates cannot be translated trivially. For example, candidate 11 curries to $((p \to q) \to r) \to ((p \to r) \to r \to s) \to q \to s$, and a lambda term with that type would have to be of the form $\lambda xyz.yEF$ where $E:p \to r$ and $F:r$. Taken separately this is easy: $F$ must be $x(\lambda u.z)$, and $E$ must be $\lambda v.F$. But the most general type of $\lambda xyz.y(\lambda v.x(\lambda u.z))(x(\lambda u.z))$ is $$((p \to q) \to r) \to ((s \to r) \to r \to t) \to q \to t$$

(which happens to be candidate 4).

My intuition is that this should rule out candidate 11, but it seems far more likely that my intuition is failing than that Ulrich and John Halleck (who has obviously given some thought to how to rule out the candidates, since he's knocked down four of them) have overlooked something so simple.

Where does my intuition fall down? Is it just undecidability which says that it's more complicated than that to show that no term has the right type? Is this a reasonable and known heuristic argument for the failure of candidate 4 (and the others for which similar arguments can be made) which no-one has managed to make rigorous?