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Proof step 14 from proof afs-cons-rename-dir-dentry (previous step)

dirs[dirs[m].entries[str]].entries = ∅, boolvar, m1 = dirs[m0].nlink, m2 = dirs[m0].nsubdirs, m3 = dirs[m0].size, new_child_inode = mkinode(dirs[m].entries[str], md, true, dirs[dirs[m].entries[str]].nlink, dirs[dirs[m].entries[str]].nsubdirs, dirs[dirs[m].entries[str]].size), new_dent = mkdentry(str, dirs[m].entries[str]), new_parent_inode = mkinode(m, md1, true, dirs[m].nlink, dirs[m].nsubdirs, dirs[m].size), old_child_inode = mkinode (dirs[m0].entries[str0], md0, true, dirs[dirs[m0].entries[str0]].nlink, dirs[dirs[m0].entries[str0]].nsubdirs, dirs[dirs[m0].entries[str0]].size), old_dent = mkdentry(str0, dirs[m0].entries[str0]), old_parent_inode = mkinode(m0, md2, boolvar, m1, m2, m3), afs-cons(dirs, files), m0 ∈ dirs, m ∈ dirs, dirs[m0].entries[str0] ∈ dirs, dirs[m].entries[str] ∈ dirs, str0 ∈ dirs[m0].entries, str ∈ dirs[m].entries, m = m0 → md1 = md2, links(m0, dirs) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ ⟪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), dirs[m].entries[str] = ROOT_INO

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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