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Proof step 5 from proof first-append (previous step)

m0 < # x, ∀ x, y. m0 ≤ # x → firstn(m0, x + y) = firstn(m0, x) ⊦ firstn(m0, x.tail + y) = firstn(m0, x.tail), x = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
x = a + y0, m0 < # x, ∀ x, y. m0 ≤ # x → firstn(m0, x + y) = firstn(m0, x) ⊦ firstn(m0, y0 + y) = firstn(m0, y0), x = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x, a, y1]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x, a, y1