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Proof step 10 from proof in-vl-append0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information