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Proof step 16 from proof sublist-prefix (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m, m0, y

Additional information

Used terms: m, m0, y