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Proof step 9 from proof get-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n ≤ # x0, ∀ n0, a1. n0 < # x0 → (x0 + a1 ')[n0] = x0[n0] ⊦ (x0 + a ')[n - 1] = (a0 ' + x0)[n], n = 0, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)