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Proof step 4 from proof restN-append (previous step)

n0 < # x, ∀ x, y. n0 ≤ # x → restn(n0, x + y) = restn(n0, x) + y ⊦ x = [], restn(n0, x.rest + y) = restn(n0, x.rest) + y

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a + y0, n0 < # x, ∀ x, y. n0 ≤ # x → restn(n0, x + y) = restn(n0, x) + y ⊦ x = [], restn(n0, y0 + y) = restn(n0, y0) + y

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, y1] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y1