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Proof step 2 from proof restn-less (previous step)

n < # x ⊦ restn(n, x) = x[n] ' + restn(n, x.rest), x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x = a + y, n < # x ⊦ restn(n, x) = x[n] ' + restn(n, y), x = []

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