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Proof step 6 from proof restn-firstn-better (previous step)

m0 < # x, n0 + m0 < # x, ∀ n0, x. n0 + m0 ≤ # x ∧ m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, x.rest)) = firstn(n0, restn(m0, x.rest)), x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
x = a + y, m0 < # x, n0 + m0 < # x, ∀ n0, x. n0 + m0 ≤ # x ∧ m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, y)) = firstn(n0, restn(m0, 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