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Proof step 2 from proof afs-cons-non0 (previous step)

entries[str] = 0, dirs[ino] = mkdir(n0, md, n, n1, entries), afs-cons(dirs, files), ino ∈ dirs, str ∈ entries ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n;