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Proof step 44 from proof afs-cons-mkdir-ino (previous step)

(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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
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, 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 ∈ entries[str, ino] ∧ (entries[str, ino][str0] = ino ∨ entries[str, ino][str0] ∈ dirs) ↔ str0 = str ∨ str0 ∈ entries ∧ entries[str0] ∈ dirs, entries = ∅, ino = 0, str ∈ entries, ino ∈ files, ino ∈ dirs, m ≠ ROOT_INO ∧ links(m, dirs) = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ entries[str, ino][str] = ino; (s)% ⊦ str ∈ entries[str, ino]; (s)% ⊦ ¬ str ∈ ∅; (s)% ⊦ str ≠ str0 → (str ∈ entries[str0, ino] ↔ str ∈ entries); (s)% ⊦ # ∅ = 0; (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)%ino ∈ dirs, ino0 ≠ 0, ¬ ino0 ∈ dirs ⊦ ino ≠ ino0 → (dirs ++ ino0)[ino] = dirs[ino]; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)