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Proof step 6 from proof sublist-first-skip (previous step)

n2 = n0 + n, n ≤ n2, n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x, n ≤ # x ⊦ firstn(n1, firstn(n0, restn(n, x))) = firstn(n1, restn(n, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n + n1 ≤ # x, n ≤ # x, n0 + n ≤ # x, n1 ≤ n0 ⊦ firstn(n1, firstn(n0, restn(n, x))) = firstn(n1, restn(n, x))

Rule argument

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat