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Proof step 7 from proof rest-sub (previous step)

# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → restn(n1, x) = firstn(n0, restn(n1, x)) ⊦ restn(n1, x.tail) = firstn(n0, restn(n1, x.tail))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
# x = n0 + n1 + 1, x = a + y, ∀ x, n0. # x = n0 + n1 → restn(n1, x) = firstn(n0, restn(n1, x)) ⊦ restn(n1, y) = firstn(n0, restn(n1, y)), x = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a, y]

Additional information

Used simplifier rules: ⊦ # [] = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a, y