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Proof step 27 from proof in-vl-in-lebrefs (previous step)

eq-entry(n, lref) ∈ q, ∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ eq-entry(n, lref) = eq-entry(m, lref0), lref = lref0, lref = none, lref0 = none, lref ∈ q.lebrefs, none ∈ q.lebrefs, ∃ n. eq-entry(n, lref) ∈ q

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 7 terms: n

Additional information

Used terms: n