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Proof step 10 from proof get-firstn (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n = m + 1, m0 ≤ # x0, n ≤ m0, n ≤ # x0, ∀ n, x0, a. n ≤ # x0 ∧ m0 ≤ # x0 + 1 ∧ n < m0 ∧ (∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0) → x0 = [] ∨ firstn(m0, a ' + x0)[n] = (a ' + x0)[n] ∨ m0 = 0, ∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0 ⊦ n = 0, x0 = [], firstn(m0, x0)[m] = (a ' + x0)[n]

Rule argument

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

Additional information

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