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Proof step 5 from proof sublist-lenrec (previous step)

(m0 + 1) + n < # x, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).first ' + firstn(n0, restn(m0, x0).rest) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ x = [], restn(m0, x.rest).first ' + firstn(n, restn(m0, x.rest).rest) = firstn(n, restn(m0, x.rest)) + x.rest[m0 + n] '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
x = a + y, (m0 + 1) + n < # x, ∀ n0, x0. m0 + n0 < # x0 → restn(m0, x0).first ' + firstn(n0, restn(m0, x0).rest) = firstn(n0, restn(m0, x0)) + x0[m0 + n0] ' ⊦ x = [], restn(m0, y).first ' + firstn(n, restn(m0, y).rest) = firstn(n, restn(m0, y)) + y[m0 + n] '

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