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Proof step 2 from proof sublist-append-01 (previous step)

# x ≤ n ⊦ firstn(m, restn(n, x + y)) = firstn(m, restn(n - # x, y))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + # x, # x ≤ n ⊦ firstn(m, restn(n, x + y)) = firstn(m, restn(n0, y))

Rule argument

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

Additional information

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