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Proof step 9 from proof diff-in (previous step)

a0 ∈ y, a ∈ x0 \ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a, y

Additional information

Used terms: a, y