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Proof step 7 from proof keys-empty (previous step)

∀ key. ¬(∃ n. key = inodekey(n) ∧ n ∈ ns) ⊦ # ns = 0, ∀ n. n ∈ ns ↔ n ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ∈ ns, ∀ key. ¬(∃ n. key = inodekey(n) ∧ n ∈ ns) ⊦ # ns = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ∅; (s)