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Proof step 9 from proof subdel-01 (previous step)

#oc(a0, (y + y0) -1l a) < #oc(a0, x), a ∈ y0, a0 ∈ x, #oc(a0, x) ≤ #oc(a0, y) + #oc(a0, y0 -1l a), ∀ a1. #oc(a1, x) ≤ #oc(a1, y) + #oc(a1, y0 -1l a) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
#oc(a0, (y + y0) -1l a) < #oc(a0, y) + #oc(a0, y0 -1l a), #oc(a0, x) ≤ #oc(a0, y) + #oc(a0, y0 -1l a), a0 ∈ x, a ∈ y0, #oc(a0, (y + y0) -1l a) < #oc(a0, x), ∀ a1. #oc(a1, x) ≤ #oc(a1, y) + #oc(a1, y0 -1l a) ⊦ #oc(a0, y) + #oc(a0, y0 -1l a) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat