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Proof step 7 from proof oc-diff-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, (a0 ' + x0) \ y) = #oc(a, x0), a ∈ y

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2