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

∀ key. key ∈ ns.keys ↔ key ∈ ∅ ⊦ # ns = 0, ns = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ key. ¬(∃ n. key = inodekey(n) ∧ n ∈ ns) ⊦ # ns = 0, ns = ∅

Rule argument

Additional information

Used simplifier rules: keys-in: ⊦ key ∈ ns.keys ↔ (∃ n. key = inodekey(n) ∧ n ∈ ns);% ⊦ ¬ key ∈ ∅; (s)