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Proof step 9 from proof sublist-prefix (previous step)

n1 = m + 1, n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ n1 = 0, x = [], firstn(n0, restn(m0, x).tail.tail) ⊑ firstn(m, restn(m0, x).tail.tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(m + 1) + m0 < # x, n0 ≤ m, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ x = [], firstn(n0, restn(m0, x).tail.tail) ⊑ firstn(m, restn(m0, x).tail.tail)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)