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Proof step 10 from proof subseteq-lebrefs-subseteq (previous step)

eq-entry(m, lref) ∈ q1, q ⊆ q1, q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ lref = none, q.lebrefs ++ lref ⊆ q1.lebrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: in-vl-in-lebrefs: ⊦ lref ∈ q0.lebrefs ↔ lref ∈ q0 ∧ lref ≠ none;%queue-in-vl: ⊦ lref ∈ q ↔ (∃ n. eq-entry(n, lref) ∈ q);% ⊦ s1 ++ lref ⊆ s2 ↔ s1 ⊆ s2 ∧ lref ∈ s2; (ws)