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Proof step 5 from proof get-append-less (previous step)

n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ ((a ' + x0) + y)[n] = (a ' + x0)[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 7:
n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ (a ' + x0) + y = [], ((a ' + x0) + y).tail[n - 1] = (a ' + x0)[n], n = 0

Rule argument

spec: <toplevel> name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1] subst: [n, x] → [n, (a ' + x0) + y] flags: <all paths>

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, (a ' + x0) + y