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Proof step 6 from proof in-prefixn-in-shorter-prefix (previous step)

a ∈ firstn(n0, x), a ∈ x, n0 ≤ n1, n1 ≤ # x, n0 ≤ # x ⊦ a ∈ firstn(n1, x)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
# y = n0, x = y + z, a ∈ x, a ∈ y, n0 ≤ # x, n1 ≤ # x, n0 ≤ n1 ⊦ a ∈ firstn(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, 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, y, z