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Proof step 3 from proof oc-diff-01 (previous step)

∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, (a0 + x0) \ y) = #oc(a, a0 + x0), a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ #oc(a, (a0 ' + x0) \ y) = #oc(a, a0 ') + #oc(a, x0), a ∈ y

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%c: ⊦ a + x = a ' + x; from specification list