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Proof step 11 from proof in-vl-in-lebrefs (previous step)

lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none, lref = lref0 ∨ lref ∈ q.lebrefs ⊦ #(q.lebrefs ++ lref0) = 0, lref0 = none, (∃ n. eq-entry(n, lref) = eq-entry(m, lref0) ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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