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Proof step 9 from proof restn-nonempty (previous step)

restn(n -1, (a ' + a0 ' + x0).tail) = [], 0 < n, n ≤ # x0 ⊦ x0 = [], ∃ n, a. restn(n, a ' + x0) = [] ∧ n < # x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
restn(n - 1, a0 ' + x0) = [], n ≤ # x0 ⊦ n = 0, x0 = [], ∃ n, a. restn(n, a ' + x0) = [] ∧ n < # 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)