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Proof step 1 from proof sublist-first-skip (conclusion)

⊦ n2 ≤ # x ∧ n + n1 ≤ n2 → sublist(n, n1, firstn(n2, x)) = sublist(n, n1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x ⊦ firstn(n1, restn(n, firstn(n2, x))) = firstn(n1, restn(n, x))

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat