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Proof step 6 from proof del1 (previous step)

#oc(a, x0) = m + 1, a ∈ x0, ∀ a, m. #oc(a, x0) = m + 1 ∧ a ∈ x0 → #oc(a, x0 -1l a) = m ⊦ a = a0, #oc(a, x0 -1l a) = m

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a, m

Additional information

Used terms: a, m