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Proof step 184 from proof afs-cons-rename-file-dentry (previous step)

# links(entries0[str], dirs0) = m11 + 1, (m = m0 ⊃ mkdir(m3, md4, # links(m0, dirs0), m7, entries1 -- str0);mkdir(m4, md6, # links(m, dirs0), m8, entries0)) = mkdir(m6, md5, m9, m10, entries), dirs0[m] = mkdir(m4, md6, # links(m, dirs0), m8, entries0), files0[entries0[str]] = mkfile(m5, md0, m11 + 1, content1), files0[entries1[str0]] = mkfile(m1, md3, # links(entries1[str0], dirs0), content), dirs0[m0] = mkdir(m3 + 1, md4, # links(m0, dirs0), m7, entries1), dirs = dirs0[m0, (m0 = m ⊃ mkdir(m6, md5, m9, m10, entries[str, entries1[str0]]);mkdir(m3, md4, # links(m0, dirs0), m7, entries1 -- str0)).meta:= md2] [m, (m = m0 ⊃ mkdir(m6, md2, m9, m10, entries[str, entries1[str0]]);mkdir(m6, md5, m9, m10, entries[str, entries1[str0]])).meta:= md1 ], files = files0[entries0[str], mkfile(m5, md0, m11, content1)], ino = inode.ino, inode = mkinode(entries0[str], md, false, # links(entries0[str], dirs0), 0, m2), afs-cons(dirs0, files0), m ∈ dirs0, str0 ∈ entries1, entries0[str] ∈ files0, entries1[str0] ∈ files0, str ∈ entries0, m0 ∈ dirs0, m = m0 → md1 = md2, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ⟪new_child_inode := getinode(entries0[str], dirs, files); old_parent_inode := (m0 ∈ dirs ⊃ mkinode(m0, dirs[m0].meta, true, dirs[m0].nlink, dirs[m0].nsubdirs, dirs[m0].size) ;mkinode(m0, files[m0].meta, false, files[m0].nlink, 0, files[m0].size) ); new_parent_inode := (m ∈ dirs ⊃ mkinode(m, dirs[m].meta, true, dirs[m].nlink, dirs[m].nsubdirs, dirs[m].size) ;mkinode(m, files[m].meta, false, files[m].nlink, 0, files[m].size) ); err := ESUCCESS; asynced := ?⟫ afs-cons(dirs, files), ino ∈ dirs, entries1[str0] ∈ dirs0, entries0[str] ∈ dirs0, ino ∈ files, m0 = m ∧ mkdentry(str, entries0[str]) = mkdentry(str0, entries1[str0])

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)%⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)