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Proof step 4 from proof restn-nonempty (previous step)

restn(n, a ' + (a0 + x0)) = [], n < #(a0 + x0), ∀ n, a. restn(n, a ' + x0) = [] ∧ n < # x0 → x0 = [] ⊦ a0 + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
restn(n, a ' + a0 ' + x0) = [], n ≤ # x0, (∃ n, a. restn(n, a ' + x0) = [] ∧ n < # x0) → x0 = [] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)