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Proof step 18 from proof afs-cons-rename-negdentry (previous step)

dirs[m] = mkdir(m6, md3, # links(m, dirs), m8, entries), files[entries0[str]] = mkfile(m1, md2, m5, content), dirs[m0] = mkdir(m7, md4, m9, m10, entries0), new_dent = negdentry(str0), new_parent_inode = mkinode(m, md, true, # links(m, dirs), m8, m6), old_child_inode = mkinode(entries0[str], md1, boolvar, m2, m3, m4), old_dent = mkdentry(str, entries0[str]), old_parent_inode = mkinode(m0, md0, true, m9, m10, m7), afs-cons(dirs, files), m ∈ dirs, m0 ∈ dirs, str ∈ entries0, m = m0 → md = md0, entries0[str] ∈ dirs ∧ m2 = dirs[entries0[str]].nlink ∧ m4 = dirs[entries0[str]].size ∧ m3 = dirs[entries0[str]].nsubdirs ∧ boolvar ∨ entries0[str] ∈ files ∧ m2 = m5 ∧ m3 = 0 ∧ ¬ boolvar, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs) ≠ ∅ ∨ m0 = ROOT_INO, entries0[str] ∈ dirs ∨ entries0[str] ∈ files ⊦ ⟪afs_rename#(; old_parent_inode, new_parent_inode, old_child_inode, new_child_inode, old_dent, new_dent, err, dirs, files; asynced)⟫ afs-cons(dirs, files), str0 ∈ entries

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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; (forward) from specification afs-base