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Proof step 7 from proof subdel-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
#oc(a0, (y + y0) -1l a) < #oc(a0, x), a ∈ y0, a0 ∈ x, ∀ a1. #oc(a1, x) ≤ #oc(a1, y) + #oc(a1, y0 -1l a) ⊦

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list