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Proof step 7 from proof sublist-prefix (previous step)

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 = [], restn(m0, x).tail.head ' + firstn(n0, restn(m0, x).tail.tail) ⊑ restn(m0, x).tail.head ' + firstn(n1 -1, restn(m0, x).tail.tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 < n1, n1 + m0 < # x, ∀ 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(n1 - 1, restn(m0, x).tail.tail)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y ⊑ x + z ↔ y ⊑ z; (s)% ⊦ n -1 = n - 1; (s)% ⊦ m < n → n ≠ 0; (s)