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Proof step 5 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 6:
∀ a0. #oc(a0, a ') = 0 ⊦

Rule argument

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%z: ⊦ 0 + m = m; from specification nat-basic2%less-1: ⊦ ¬ n < 0; from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%size-01: ⊦ # [] = 0; from specification list-data