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Proof step 34 from proof afs-cons-create-ino (previous step)

dirs[m] = mkdir(# entries0, md3, # links(m, dirs), # subdirs(m, dirs), entries0), files[m] = mkfile(m1, md1, m3, content), (files ++ ino)[m] = mkfile(m4, md2, m5, content0), afs-cons(dirs, files), m ∈ dirs, ROOT_INO ∈ dirs, # links(m, dirs) ≤ 1, m ≠ ROOT_INO ∧ links(m, dirs) = ∅ → entries0 = ∅, name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] ∈ files, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ ino = 0, name = str, str ∈ entries0, ino ∈ dirs, ino ∈ files, (m = ROOT_INO ∨ links(m, dirs) ≠ ∅) ∧ (name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] = ino ∨ entries0[name] ∈ files)

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: m = ROOT_INO

Additional information

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