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Proof step 6 from proof subseteq-lebrefs-subseteq (previous step)

eqent = eq-entry(m, lref), eqent ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ q.lebrefs ∪ (lref ≠ none ⊃ ∅ ++ lref;∅) ⊆ q1.lebrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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