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Proof step 4 from proof prefixn-split-sublist (previous step)

n0 + n1 ≤ # x ⊦ firstn(n0 + n1, x) = firstn(n0, x) + sublist(n0, n1, x)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
# y = n0 + n1, x = y + z, n0 + n1 ≤ # x ⊦ y = firstn(n0, x) + sublist(n0, n1, x)

Rule argument

spec: list-del name: elim-firstn-restn seq: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n) subst: [x, n, y, z] → [x, n0 + n1, y, z]

Additional information

Used simplifier rules: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n); (elim) Used terms: x, n0 + n1, y, z