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Proof step 7 from proof firstn-firstn (previous step)

m0 < n, m0 ≤ # y, n ≤ 1 + # y, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(n, a ' + y).head ' + firstn(m0, firstn(n, a ' + y).tail) = a ' + firstn(m0, y)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
firstn(n, a ' + y) = a0 + y0, m0 < n, n ≤ 1 + # y, m0 ≤ # y, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ a0 ' + firstn(m0, y0) = a ' + firstn(m0, y), firstn(n, a ' + y) = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [firstn(n, a ' + y), a0, y0]

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;% ⊦ m < n → n ≠ 0; (s)%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: firstn(n, a ' + y), a0, y0