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Proof step 13 from proof subsize (previous step)

∃ y0, z. y = y0 + (a + z), #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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
#oc(a, x0) ≤ #oc(a, y0) + #oc(a, z), # y0 + # z < # x0, ∀ 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;%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list