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Proof step 6 from proof del1-01 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a, b

Additional information

Used terms: a, b