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Proof step 7 from proof rev-disjoint-01 (previous step)

disj(rev(x), y0), ∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ disj(x, y0), a ∈ x

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: x

Additional information

Used terms: x