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Proof step 8 from proof get-butlast (previous step)

(a0 ' + x0) + a ' ≠ [] ∧ n ≠ 0, n ≤ # x0, ∀ n0, a1. n0 < # x0 → (x0 + a1 ')[n0] = x0[n0] ⊦ x0 = [], ((a0 ' + x0) + a ').rest[n - 1] = (a0 ' + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ # x0, ∀ n0, a1. n0 < # x0 → (x0 + a1 ')[n0] = x0[n0] ⊦ n = 0, x0 = [], (x0 + a ')[n - 1] = (a0 ' + 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