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Proof step 12 from proof get-firstn (previous step)

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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m < # x0, m < 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] = x0[m], x0 = []

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];% ⊦ m < n → n ≠ 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)