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Proof step 7 from proof firstn-append (previous step)

y0 = firstn(n0, y0) + a ' + y, ∀ a0, y0, a, y. a0 ' + y0 = firstn(n0, a0 ' + y0) + a ' + y ∧ (a0 ∈ firstn(n0, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y) → a0 ' + firstn(n0, y0) = firstn(n0, a0 ' + y0) + a ' ⊦ y0.head ' + firstn(n0, y0.tail) = firstn(n0, y0) + a '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
y0 = firstn(n0, y0) + a ' + y, y0 = a1 + y1, ∀ a0, y0, a, y. a0 ' + y0 = firstn(n0, a0 ' + y0) + a ' + y ∧ (a0 ∈ firstn(n0, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y) → a0 ' + firstn(n0, y0) = firstn(n0, a0 ' + y0) + a ' ⊦ a1 ' + firstn(n0, y1) = firstn(n0, y0) + a ', y0 = []

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [y0, a1, y1]

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: y0, a1, y1