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Proof step 3 from proof in-vl-in-lebrefs (previous step)

∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ lref ∈ (eqent + q).lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ eqent + q) ∧ lref ≠ none

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ lref ∈ q.lebrefs ∪ (eqent.lebref ≠ none ⊃ ∅ ++ eqent.lebref;∅) ↔ (∃ n. eq-entry(n, lref) = eqent ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Rule argument

Additional information

Used simplifier rules: lebrefs-rec: ⊦ (eqent ' + q).lebrefs = q.lebrefs ∪ (eqent.lebref ≠ none ⊃ {eqent.lebref} ;∅);% ⊦ eqent ∈ eqent0 ' + q ↔ eqent = eqent0 ∨ eqent ∈ q; (s)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ {lref} = ∅ ++ lref; (s)% ⊦ eqent + q = eqent ' + q; (s)