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Proof step 6 from proof in (previous step)

a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ a = a0, a ∈ x0 -l a0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, a0

Additional information

Used terms: a, a0