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Proof step 10 from proof oc-diff-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, x0 \ y) = #oc(a, x0), a ∈ y, a0 ∈ y

Rule argument

Additional information

Used simplifier rules: oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y); from specification list-set