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Proof step 2 from proof afs-cons-links-empty (previous step)

afs-cons(dirs, files) ⊦ links(ino, dirs) = ∅ ↔ (ino ∈ dirs → dirs[ino].nlink = 0) ∧ (ino ∈ files → files[ino].nlink = 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information