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Proof step 28 from proof in-vl-in-lebrefs (previous step)

lref0 = none, ∀ 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 ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
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

Rule argument

Additional information

Used simplifier rules: ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)% ⊦ ¬ lref ∈ ∅; (s)