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

a ∈ x0 -1l b, ∀ a, b. a ∈ x0 -1l b → a ∈ x0 ⊦ b = a0, a = a0, a ∈ x0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, b

Additional information

Used terms: a, b