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Proof step 5 from proof in-vl-in-lebrefs (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ lref. lref ∈ q.lebrefs ↔ (∃ n. eq-entry(n, lref) ∈ q) ∧ lref ≠ none ⊦ lref ∈ q.lebrefs ∪ (lref0 ≠ none ⊃ ∅ ++ lref0;∅) ↔ (∃ n. eq-entry(n, lref) = eq-entry(m, lref0) ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Proof step 5 from proof in-vl-in-lebrefs (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information