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Proof step 10 from proof orderedls-all (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 12:
m < n, a < a0, ordered<(a0 ' + y), m ≤ # y, n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ (a ' + a0 ' + y)[m] < (a ' + a0 ' + y)[n], a ' + a0 ' + y ≠ [] ∧ m ≠ 0
Proof step 22:
m < n, a < a0, ordered<(a0 ' + y), m ≤ # y, n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ a ' + a0 ' + y = [], m = 0, (a ' + a0 ' + y).tail[m - 1] < (a ' + a0 ' + y)[n]

Rule argument

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

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m, a ' + a0 ' + y