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Proof step 8 from proof subseteq-lebrefs-subseteq (previous step)

eq-entry(m, lref) ∈ q1, q ⊆ q1, ∀ 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

Proof step 9:
eq-entry(m, lref) ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ lref = none, q.lebrefs ++ lref ⊆ q1.lebrefs

Rule argument

Additional information

Used simplifier rules: unc: ⊦ s1 ∪ s2 = s2 ∪ s1; from specification set-union[lebrefset]%una: ⊦ (s2 ∪ s3) ∪ s4 = s2 ∪ s3 ∪ s4; from specification set-union[lebrefset]% ⊦ s0 ∪ ∅ = s0; (s)% ⊦ (s1 ++ lref) ∪ s2 = (s1 ∪ s2) ++ lref; (s)