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Proof step 15 from proof subrec (previous step)

a ∈ y ∧ (∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a)) ⊦ ∀ a0. #oc(a0, a ') + #oc(a0, x) ≤ #oc(a0, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
a ∈ y, #oc(a0, y) < #oc(a0, a ') + #oc(a0, x), ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦ #oc(a0, a ') + #oc(a0, x) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat