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Proof step 6 from proof ad (previous step)

a ∈ y ⊦ #oc(a1, (x + y) -1l a) ≤ #oc(a1, x) + #oc(a1, y -1l a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a1 ∈ (x + y) -1l a, a ∈ y, #oc(a1, x) + #oc(a1, y -1l a) < #oc(a1, (x + y) -1l a), a1 ∈ x ∨ a1 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-01-01: ⊦ a ∈ x -1l b → a ∈ x; (forward) from specification list-del%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat