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Proof step 6 from proof subsize (previous step)

x0 = [], # y ≤ # x0, (∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))) → x0 = [], ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a0. #oc(a0, a ') = 0 ⊦

Rule argument

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0;% ⊦ ¬ n < 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 0 + m = m; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ # [] = 0; (s)