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Proof step 26 from proof sublist-move (previous step)

n + m ≤ m0, m1 + m0 ≤ # x0, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n, firstn(m0, restn(m1, x0)))) = firstn(m, restn(m1 + n, x0))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, m, m0, m1

Additional information

Used terms: n, m, m0, m1