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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

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