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Proof step 41 from proof afs-cons-mkdir-ino (previous step)

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), ino = 0, str ∈ entries, ino ∈ files, ino ∈ dirs, ∀ str0. str0 ∈ subdirs(m, dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino) ↔ str0 ∈ subdirs(m, dirs) ++ str

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
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 ∈ (dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino)[m].entries ∧ ( (dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino)[m].entries[str0] = ino ∨ (dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino)[m].entries[str0] ∈ dirs) ↔ str0 = str ∨ str0 ∈ entries ∧ entries[str0] ∈ dirs, ino = 0, str ∈ entries, ino ∈ files, ino ∈ dirs

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ str ∈ strings ++ str0 ↔ str = str0 ∨ str ∈ strings; (s)%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)%⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs; (s)