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Proof step 4 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 5:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1]
Proof step 6:
(a ' + x0) + y ≠ [] ∧ n ≠ 0, n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ ((a ' + x0) + y).rest[n - 1] = (a ' + x0)[n]

Rule argument

spec: list-del inst: name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1] subst: [n, x] → [n, (a ' + x0) + y] flags: rewrite, current

Additional information

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