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Proof step 16 from proof subsize (previous step)

# y0 + # z < # x0, #oc(a, x0) ≤ #oc(a, y0) + #oc(a, z), ∀ a0. #oc(a0, x0) ≤ #oc(a0, y0) + #oc(a0, z) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y)), #(y0 + z) < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y0 + z))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
#oc(a0, y0) + #oc(a0, z) < #oc(a0, x0), # y0 + # z < # x0, #oc(a, x0) ≤ #oc(a, y0) + #oc(a, z), ∀ a0. #oc(a0, x0) ≤ #oc(a0, y0) + #oc(a0, z) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)