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Proof step 11 from proof get-firstn (previous step)

m0 ≤ # x0, n ≤ m0, n ≤ # x0, ∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0], ∀ n, x0, a. n < m0 ∧ n ≤ # x0 ∧ m0 ≤ 1 + # x0 ∧ (∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0]) → firstn(m0, a ' + x0)[n] = (a ' + x0)[n] ∨ x0 = [] ⊦ firstn(m0, x0)[n - 1] = (a ' + x0)[n], n = 0, x0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
n = m + 1, n ≤ # x0, n ≤ m0, m0 ≤ # x0, ∀ n, x0, a. n < m0 ∧ n ≤ # x0 ∧ m0 ≤ 1 + # x0 ∧ (∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0]) → firstn(m0, a ' + x0)[n] = (a ' + x0)[n] ∨ x0 = [], ∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0] ⊦ firstn(m0, x0)[m] = (a ' + x0)[n], n = 0, x0 = []

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m1