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Proof step 6 from proof in-vl-append0 (previous step)

⊦ (∃ n0. eq-entry(n0, none) ∈ q ∨ eq-entry(n0, none) = eq-entry(n, lref0)) ↔ (∃ n. eq-entry(n, none) ∈ q) ∨ none = lref0, none = lref0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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