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Proof step 1 from proof sublist-move (conclusion)

⊦ n + m ≤ m0 ∧ n0 + m0 ≤ # x → sublist(n, m, sublist(n0, m0, x)) = sublist(n0 + n, m, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n + m ≤ m0, n0 + m0 ≤ # x ⊦ firstn(m, restn(n, firstn(m0, restn(n0, x)))) = firstn(m, restn(n0 + n, x))

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));