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Proof step 43 from proof afs-cons-mkdir-ino (previous step)

files[m] = mkfile(m1, md1, m3, content), dirs[m] = mkdir(# entries, md0, # links(m, dirs), # subdirs(m, dirs), entries), (dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino)[m] = mkdir(m0, md, m2, m4, entries0), afs-cons(dirs, files), m ∈ dirs, ROOT_INO ∈ dirs, # links(m, dirs) ≤ 1, m ≠ ROOT_INO ∧ links(m, dirs) = ∅ → entries = ∅, name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ # subdirs(m, dirs) + 1 = # subdirs(m, dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino), str0 ∈ entries0 ∧ (entries0[str0] = ino ∨ entries0[str0] ∈ dirs) ↔ str0 = str ∨ str0 ∈ entries ∧ entries[str0] ∈ dirs, ino = 0, ino ∈ dirs, ino ∈ files, str ∈ entries

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 44:
(dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino)[m] = mkdir(m0, md, m2, m4, entries0), dirs[m] = mkdir(# entries, md0, # links(m, dirs), # subdirs(m, dirs), entries), files[m] = mkfile(m1, md1, m3, content), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, # links(m, dirs) ≤ 1, name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, m ≠ ROOT_INO ∧ links(m, dirs) = ∅ → entries = ∅, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ # subdirs(m, dirs) + 1 = # subdirs(m, dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino), str0 ∈ entries0 ∧ (entries0[str0] = ino ∨ entries0[str0] ∈ dirs) ↔ str0 = str ∨ str0 ∈ entries ∧ entries[str0] ∈ dirs, ino = 0, m = ino, str ∈ entries, ino ∈ files, ino ∈ dirs

Rule argument

fma: m = ino

Additional information

Used formulas: m = ino