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Proof step 28 from proof afs-cons-inode-sync (previous step)

files[ino] = mkfile(m0, md0, m1, content), dirs[ino] = mkdir(m2, md1, m3, m4, entries), mkdir(m2, md1, m3, m4, entries) = mkdir(m5, md2, m6, m7, entries0), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, ino ∈ dirs → m3 = # links(ino, dirs) ∧ # links(ino, dirs) ≤ 1 ∧ m2 = # entries ∧ m4 = # subdirs(ino, dirs), ino ∈ dirs ∧ name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, ino ∈ files → ¬ ino ∈ dirs ∧ m1 = # links(ino, dirs), ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅ → entries = ∅ ⊦ ino = m, (ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs[m, dirs[m].meta:= md]) = ∅ → entries0 = ∅) ∧ ( ino ∈ dirs → m6 = # links(ino, dirs[m, dirs[m].meta:= md]) ∧ # links(ino, dirs[m, dirs[m].meta:= md]) ≤ 1 ∧ m5 = # entries0 ∧ m7 = # subdirs(ino, dirs[m, dirs[m].meta:= md])) ∧ (ino ∈ files → ¬ ino ∈ dirs ∧ m1 = # links(ino, dirs[m, dirs[m].meta:= md])) ∧ (ino ∈ dirs ∧ name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] ∈ files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
files[ino] = mkfile(m0, md0, m1, content), dirs[ino] = mkdir(m5, md1, m6, m7, entries0), afs-cons(dirs, files), ROOT_INO ∈ dirs, m ∈ dirs, ino ∈ dirs → m6 = # links(ino, dirs) ∧ # links(ino, dirs) ≤ 1 ∧ m5 = # entries0 ∧ m7 = # subdirs(ino, dirs), ino ∈ dirs ∧ name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] ∈ files, ino ∈ files → ¬ ino ∈ dirs ∧ m1 = # links(ino, dirs), ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅ → entries0 = ∅ ⊦ ino = m, (ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs[m, dirs[m].meta:= md]) = ∅ → entries0 = ∅) ∧ ( ino ∈ dirs → m6 = # links(ino, dirs[m, dirs[m].meta:= md]) ∧ # links(ino, dirs[m, dirs[m].meta:= md]) ≤ 1 ∧ m5 = # entries0 ∧ m7 = # subdirs(ino, dirs)) ∧ (ino ∈ files → ¬ ino ∈ dirs ∧ m1 = # links(ino, dirs[m, dirs[m].meta:= md]))

Rule argument

Additional information

Used simplifier rules: ⊦ ino0 ≠ ino1 → subdirs(ino0, dirs[ino1, ad]) = subdirs(ino0, dirs); (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)