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Proof step 6 from proof restn-less (previous step)

n0 < # y, ∀ y, a. n0 ≤ # y → restn(n0, a ' + y) = (a ' + y)[n0] ' + restn(n0, y) ⊦ restn(n0, y) = y[n0] ' + restn(n0, y.rest), y = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
y = a0 + y0, n0 < # y, ∀ y, a. n0 ≤ # y → restn(n0, a ' + y) = (a ' + y)[n0] ' + restn(n0, y) ⊦ restn(n0, y) = y[n0] ' + restn(n0, y0), 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, y0] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: y, a0, y0