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Proof step 3 from proof sublist-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n1 + m0 < # x, n0 < n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], n1 = 0, restn(m0, x).rest.first ' + firstn(n0, restn(m0, x).rest.rest) ⊑ firstn(n1, restn(m0, x).rest)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2