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Proof step 45 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, 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) = ∅

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: str0 = str

Additional information

Used simplifier rules: ⊦ str ≠ str0 → (str ∈ entries[str0, ino] ↔ str ∈ entries); (s)%⊦ str ∈ entries ∧ ¬ str0 ∈ entries → entries[str0, ino][str] = entries[str]; (s)%⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries → entries[str] ≠ 0; (s)%⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)% ⊦ str ∈ entries[str, ino]; (s)% ⊦ entries[str, ino][str] = ino; (s) Used formulas: str0 = str