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Proof step 14 from proof in-vl-in-lebrefs (previous step)

lref ∈ q.lebrefs, ∃ n. eq-entry(n, lref) ∈ q, lref = lref0 ∨ lref ∈ q.lebrefs ⊦ #(q.lebrefs ++ lref0) = 0, lref = none, lref0 = none, (∃ n. eq-entry(n, lref) = eq-entry(m, lref0) ∨ eq-entry(n, lref) ∈ q) ∧ lref ≠ none

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
lref ∈ q.lebrefs, eq-entry(n, lref) ∈ q ⊦ #(q.lebrefs ++ lref0) = 0, # q.lebrefs = 0, q.lebrefs = ∅, lref = none, lref0 = none, ∃ n. eq-entry(n, lref) = eq-entry(m, lref0) ∨ eq-entry(n, lref) ∈ q

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s0 ≠ ∅ → # s0 ≠ 0; (forward) from specification set-basic[lebrefset]%ne-01: ⊦ lref ∈ s0 → s0 ≠ ∅; (forward) from specification set-basic[lebrefset]