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Proof step 11 from proof del1-01 (previous step)

∀ a, b. #oc(a, x0 -1l b) = #oc(a, x0) ∨ a = b ⊦ #oc(a, a0 ' + (x0 -1l b)) = #oc(a, x0), a = b, a0 = a, b = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ a, b. #oc(a, x0 -1l b) = #oc(a, x0) ∨ a = b ⊦ #oc(a, x0 -1l b) = #oc(a, x0), a = b, a0 = a, b = a0

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