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Proof step 5 from proof firstn-firstn (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
x = a + y, m0 < n, m0 < # x, n ≤ # x, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(n, x).head ' + firstn(m0, firstn(n, x).tail) = a ' + firstn(m0, 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: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a, y