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Proof step 8 from proof in-vl-append0 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
eq-entry(n0, none) ∈ q ⊦ eq-entry(n0, none) = eq-entry(n, lref0), none = lref0, ∃ n. eq-entry(n, none) ∈ q

Rule argument

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)