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Proof step 10 from proof delapp (previous step)

x = a1 + y0, #oc(a1, x) + #oc(a1, y) ≤ #oc(a1, y -1l a), ∀ a0. #oc(a0, x) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, x) + #oc(a0, y) ⊦ x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
#oc(a1, y0) + #oc(a1, y) < #oc(a1, y -1l a), ∀ a0. (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ (#oc(a0, a1 ') + #oc(a0, y0)) + #oc(a0, y) ⊦

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%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% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)