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Proof step 10 from proof subsize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
#oc(a, x0) < #oc(a, y), # y ≤ # x0, ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)