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Proof step 6 from proof get-append-less (previous step)

(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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list