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Proof step 5 from proof sublist-first-skip (previous step)

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))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
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))

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n2, n, n0] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, n, n0