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Proof step 4 from proof sublist-last (previous step)

m0 + n0 + 1 < # x, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ x = [], (restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest)).last = x.rest[m0 + n0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
x = a + y, m0 + n0 + 1 < # x, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ x = [], (restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest)).last = y[m0 + 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