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Proof step 4 from proof sublist-first-skip (previous step)

n ≤ n2 ∧ n2 ≤ # x, n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x ⊦ firstn(n1, firstn(n2 - n, restn(n, x))) = firstn(n1, restn(n, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ n2, n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x, n ≤ # x ⊦ firstn(n1, firstn(n2 - n, restn(n, x))) = firstn(n1, restn(n, x))

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat