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Proof step 6 from proof inodes-put (
previous step
)
ks.inodes?, ∀ key0. key0 = key ∨ key0 ∈ ks → key0.inode? ⊦ 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