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Proof step 6 from proof restn-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
restn(n - 1, x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = 1 + # x0, n = 0

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)