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Proof step 9 from proof firstn-last (previous step)

m < # y, ∀ y, a. m ≤ # y → (a ' + firstn(m, y)).last = (a ' + y)[m] ⊦ (y.head ' + firstn(m, y.tail)).last = y[m], y = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
y = a0 + y0, m < # y, ∀ y, a. m ≤ # y → (a ' + firstn(m, y)).last = (a ' + y)[m] ⊦ (a0 ' + firstn(m, y0)).last = y[m], y = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [y, a0, y1]

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y, a0, y1