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

ns.keys.inodes?, key ∈ ns.keys ⊦ # ns.keys = 0, ns.keys = ∅, ∃ n. key = inodekey(n) ∧ n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information