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Proof step 9 from proof get-firstn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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]

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%e-01: ⊦ a ' + x ≠ []; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2