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Proof step 28 from proof new-inode-exists (previous step)

inodekey(ino0) ∈ ks0, ino ≤ ino0, ∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ key) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ # ks0 = 0, ks0 = ∅, key.inode?, key ∈ ks0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ino0

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: ino0