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Proof step 50 from proof afs-cons-mkdir-ino (previous step)

dirs[ino0] = mkdir(# entries1, md2, # links(ino0, dirs), # subdirs(ino0, dirs), entries1), files[ino0] = mkfile(m1, md1, m3, content), dirs[m] = mkdir(m0, md0, # links(m, dirs), m2, entries), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, ino0 ∈ dirs, # links(ino0, dirs) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs ∨ entries1[name] ∈ files, ino0 ≠ ROOT_INO ∧ links(ino0, dirs) = ∅ → entries1 = ∅, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ # subdirs(ino0, dirs) = # subdirs(ino0, dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino), ino = 0, ino0 = m, str ∈ entries, ino ∈ files, ino ∈ dirs, ∀ str0. str0 ∈ subdirs(ino0, dirs) ↔ str0 ∈ subdirs(ino0, dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
dirs[ino0] = mkdir(# entries1, md2, # links(ino0, dirs), # subdirs(ino0, dirs), entries1), files[ino0] = mkfile(m1, md1, m3, content), dirs[m] = mkdir(m0, md0, # links(m, dirs), m2, entries), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, ino0 ∈ dirs, # links(ino0, dirs) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs ∨ entries1[name] ∈ files, ino0 ≠ ROOT_INO ∧ links(ino0, dirs) = ∅ → entries1 = ∅, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ # subdirs(ino0, dirs) = # subdirs(ino0, dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino), str0 ∈ entries1 ∧ entries1[str0] ∈ dirs ↔ str0 ∈ (dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino)[ino0].entries ∧ ( (dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino)[ino0].entries[str0] = ino ∨ (dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino)[ino0].entries[str0] ∈ dirs), ino = 0, ino0 = m, str ∈ entries, ino ∈ files, ino ∈ dirs

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (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)