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Proof step 8 from proof firstn-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
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 inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [y, a0, y1] flags:

Additional information

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