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

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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2