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Proof step 6 from proof ocappend (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a, y. #oc(a, x0 + y) = #oc(a, x0) + #oc(a, y) ⊦ a0 = a, #oc(a, x0 + y) = #oc(a, x0) + #oc(a, 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