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Proof step 21 from proof in-vl-in-lebrefs (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

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