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Proof step 22 from proof inv-put-inode (previous step)

files[ino].nlink = 0, err = ESUCCESS, ino = inode.ino, inode = mkinode(m, md, boolvar, m0, m1, m2), afs-cons(dirs, files -- ino), afs-cons(dirs, files), of-cons(of, dirs, files), m ∈ dirs ∧ m0 = dirs[m].nlink ∧ m2 = dirs[m].size ∧ m1 = dirs[m].nsubdirs ∧ boolvar ∨ m ∈ files ∧ m0 = files[m].nlink ∧ m1 = 0 ∧ ¬ boolvar ⊦ m = ROOT_INO, is-open(m, of), of-cons(of, dirs, files -- ino), ino ∈ dirs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ of-cons(of0, dirs0, files0) → (of-cons(of1, dirs1, files1) ↔ (∀ n. (n ∈ of0 → of0[n].ino ∈ files0) → n ∈ of1 → of1[n].ino ∈ files1))
Proof step 24:
files[of[n].ino].nlink = 0, afs-cons(dirs, files -- of[n].ino), afs-cons(dirs, files), of-cons(of, dirs, files), of[n].ino ∈ files, n ∈ of ⊦ of[n].ino = ROOT_INO, is-open(of[n].ino, of), of[n].ino ∈ dirs

Rule argument

Additional information

Used simplifier rules: of-cons-reduce: ⊦ of-cons(of0, dirs0, files0) → ( of-cons(of1, dirs1, files1) ↔ (∀ n. (n ∈ of0 → of0[n].ino ∈ files0) → n ∈ of1 → of1[n].ino ∈ files1));% ⊦ ino ∈ files -- ino0 ↔ ino ≠ ino0 ∧ ino ∈ files; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)