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Proof step 4 from proof refs-empty (previous step)

⊦ # ∅ .refs = 0, ∀ r. r ∈ ∅ .refs ↔ r ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r ∈ ∅ .refs ⊦ # ∅ .refs = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ ∅])]]