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Proof step 36 from proof afs-cons-mkdir-ino (previous step)

dirs[ino0] = mkdir(# entries0, md2, # links(ino0, dirs), # subdirs(ino0, dirs), entries0), files[ino0] = mkfile(m1, md1, m3, content), dirs[m] = mkdir(m0, md0, # links(m, dirs), m2, entries), ino0 = m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino]) = mkdir(m7, md3, m8, m9, entries1), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, ino0 ∈ dirs, # links(ino0, dirs) ≤ 1, name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] ∈ files, ino0 ≠ ROOT_INO ∧ links(ino0, dirs) = ∅ → entries0 = ∅, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ ino = 0, str ∈ entries, ino ∈ files, ino ∈ dirs, (ino0 ≠ ROOT_INO ∧ links(ino0, dirs) = ∅ → entries1 = ∅) ∧ m8 = # links(ino0, dirs) ∧ m7 = # entries1 ∧ m9 = # subdirs(ino0, dirs[m, mkdir(m0 + 1, md0, # links(m, dirs), m2 + 1, entries[str, ino])] ++ ino) ∧ (name ∈ entries1 → entries1[name] = ino ∨ entries1[name] ∈ dirs ∨ entries1[name] ∈ files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
files[m] = mkfile(m1, md1, m3, content), dirs[m] = mkdir(# entries, md0, # links(m, dirs), # subdirs(m, dirs), entries), afs-cons(dirs, files), m ∈ dirs, ROOT_INO ∈ dirs, # links(m, dirs) ≤ 1, m ≠ ROOT_INO ∧ links(m, dirs) = ∅ → entries = ∅, name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ ino = 0, ino ∈ dirs, ino ∈ files, str ∈ entries, (m = ROOT_INO ∨ links(m, dirs) ≠ ∅) ∧ # subdirs(m, dirs) + 1 = # subdirs(m, dirs[m, mkdir(# entries + 1, md0, # links(m, dirs), # subdirs(m, dirs) + 1, entries[str, ino])] ++ ino) ∧ (name ∈ entries[str, ino] → entries[str, ino][name] = ino ∨ entries[str, ino][name] ∈ dirs ∨ entries[str, ino][name] ∈ files)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ ¬ str ∈ entries → # entries[str, ino] = # entries + 1; (s)% ⊦ entries[str, ino] ≠ ∅; (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)