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Proof step 8 from proof restn-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).last = x0.last ⊦ restn(n - 1, x0).last = x0.last, n = 0, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < n ↔ n ≠ 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ n -1 = n - 1; (s)