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Proof step 31 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, none), lref = none, lref ∈ q.lebrefs, none ∈ q.lebrefs

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 32:
lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none, eq-entry(n, lref) ∈ q, ∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ eq-entry(n, lref) = eq-entry(m, none), lref = none, none ∈ q.lebrefs, lref ∈ q.lebrefs

Rule argument

pos: left 2 terms: lref

Additional information

Used terms: lref