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Proof step 30 from proof sublist-move (previous step)

n + m ≤ m0, m1 + m0 ≤ # x0, ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → 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 obtained the following new goals

Proof step 31:
m1 + m0 ≤ # x0, n + m ≤ m0, n + m ≤ # x0 ∧ m0 + m1 ≤ m → firstn(m1, restn(m0, firstn(m, restn(n, x0)))) = firstn(m1, restn(n + m0, x0)), ∀ n0, m0, n, m. n0 + m0 ≤ # x0 ∧ n + m ≤ m0 → 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))

Rule argument

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

Additional information

Used terms: n, m, m0, m1