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Proof step 5 from proof inodes-in (previous step)

key ∈ ks, ∀ key. key ∈ ks → key.inode? ⊦ # ks = 0, ks = ∅, key.inode?

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: key

Additional information

Used terms: key