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Proof step 13 from proof sublist-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n0 ≤ m, (m + 1) + m0 < # x, m + m0 < # x, firstn(n0, restn(m0, x.rest)) ⊑ firstn(m, restn(m0, x.rest)), ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).rest) ⊑ firstn(n1, restn(m0, x).rest) ⊦ x = [], firstn(n0, restn(m0, x.rest).rest) ⊑ firstn(m, restn(m0, x.rest).rest)

Rule argument

Additional information

Used simplifier rules: restn-apprec: ⊦ restn(m, restn(n, x)) = restn(m + n, x);%restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2