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Proof step 4 from proof get-restn (previous step)

m + n0 + 1 < # x, ∀ m, x. m + n0 < # x → restn(n0, x)[m] = x[m + n0] ∨ x = [] ⊦ x = [], restn(n0, x.rest)[m] = x.rest[m + n0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a + y, m + n0 + 1 < # x, ∀ m, x. m + n0 < # x → restn(n0, x)[m] = x[m + n0] ∨ x = [] ⊦ x = [], restn(n0, y)[m] = y[m + n0]

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y