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Proof step 25 from proof orderedls-all (previous step)

m1 + 1 < n, a < a0, m1 < # y, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], (a0 ' + y)[m1] < (a ' + a0 ' + y)[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 27:
m1 < # y, a < a0, m1 + 1 < n, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a ' + a0 ' + y = [], n = 0, y = [], (a0 ' + y)[m1] < (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 simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, a ' + a0 ' + y