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

n1 + m0 < # x, n ≤ n1 ⊦ x = [], firstn(n, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n1 + m0 < # x, n0 +1 ≤ n1, ∀ n1, m0, x. n1 + m0 < # x ∧ n0 ≤ n1 → x = [] ∨ firstn(n0, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail) ⊦ x = [], firstn(n0 +1, restn(m0, x).tail) ⊑ firstn(n1, restn(m0, x).tail)

Rule argument

indvar: n

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)% ⊦ [] ⊑ x; (s)