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Proof step 14 from proof subsize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# 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);%one: ⊦ #oc(a, a ') = 1;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%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)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)