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Proof step 5 from proof restn-nonempty (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
restn(n, a ' + a0 ' + x0) = [], n ≤ # x0, (∃ n, a. restn(n, a ' + x0) = [] ∧ n < # x0) → x0 = [] ⊦ x0 = []

Rule argument

fma: x0 = []

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s) Used formulas: x0 = []