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Proof step 7 from proof subseteq-lebrefs-subseteq (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
eq-entry(m, lref) ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ lref = none, q.lebrefs ∪ (∅ ++ lref) ⊆ q1.lebrefs
Proof step 12:
lref = none, eq-entry(m, lref) ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ q.lebrefs ∪ ∅ ⊆ q1.lebrefs

Rule argument

pos: right 1

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