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Proof step 2 from proof afs-cons-nlink0-empty (previous step)

dirs[ino] = mkdir(n0, md, n, n1, entries), links(ino, dirs) = ∅, afs-cons(dirs, files), ino ∈ dirs ⊦ entries = ∅, ino = ROOT_INO

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# links(ino, dirs) = 0, dirs[ino] = mkdir(n0, md, 0, n1, entries), afs-cons(dirs, files), ino ∈ dirs ⊦ entries = ∅, ino = ROOT_INO

Rule argument

Additional information

Used simplifier rules: afs-cons-links-empty: ⊦ afs-cons(dirs, files) → ( links(ino, dirs) = ∅ ↔ (ino ∈ dirs → dirs[ino].nlink = 0) ∧ (ino ∈ files → files[ino].nlink = 0));%afs-cons-notfile: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files;%afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n;%size-zero-02: ⊦ # ls = 0 → ls = ∅; (forward) from specification set-basic[alink-set]% ⊦ mkdir(m, md, m0, m1, entries).nlink = m0; (s)