How to lift lemmas

Here is a sample datatype with a deterministic relation.

datatype ty1 = A | B | C ty1 | D ty1

inductive rel1 where
  "rel1 A (C B)"
| "rel1 (C B) (D A)"

lemma rel1_det:
  "rel1 x y ⟹ rel1 x z ⟹ y = z"
  by (elim rel1.cases; auto)

I'm trying to lift the lemma for the following type:

datatype 'a ty2 = E 'a | F 'a

abbreviation "rel2 ≡ rel_ty2 rel1"

lemma rel2_det:
  "rel2 x y ⟹ rel2 x z ⟹ y = z"
  apply (cases x; cases y; auto)
  apply (metis rel1_det right_uniqueD right_uniqueI ty2.rel_intros(1) ty2.right_unique_rel)
  by (metis rel1_det right_uniqueD right_uniqueI ty2.rel_intros(2) ty2.right_unique_rel)

But the proof is very ugly. I guess it could be simplified using a transfer method, a lifting package or something else. Could you suggest how to use it?

Solution

I don't know about using the transfer package to prove this; however, it is easy to prove if you use the predicate right_unique from the library and the rules for it that the datatype package gives you for free:

lemma right_unique_rel1: "right_unique rel1"
  by (auto simp: right_unique_def elim: rel1.cases)

lemma right_unique_rel2: "right_unique rel2"
  by (intro ty2.right_unique_rel right_unique_rel1)