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Proof step 13 from proof orderedls-all (previous step)

a < a0, (a ' + a0 ' + y)[n] < a0, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a ' + a0 ' + y = [], n = 0, a < (a ' + a0 ' + y)[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 15:
(a ' + a0 ' + y).tail[n - 1] < a0, a < a0, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a ' + a0 ' + y = [], n = 0, a < (a ' + a0 ' + y).tail[n - 1]

Rule argument

spec: list-del name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1] subst: [n, x] → [n, a ' + a0 ' + y] flags: <all paths>

Additional information

Used terms: n, a ' + a0 ' + y