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Proof step 242 from proof afs-cons-rename-dir-dentry (previous step)

# links(entries[str], dirs0) = m12 + 1, (ino = m ⊃ mkdir(m8, md1, # links(m, dirs0), m11, entries[str, entries2[str0]]) ;(ino = m0 ⊃ mkdir(m1, md2, # links(m0, dirs0), m2, entries2 -- str0) ;mkdir(# entries4, md9, # links(ino, dirs0), # subdirs(ino, dirs0), entries4) ) ) = mkdir(m20, md10, m21, m22, entries5), dirs0[m0] = mkdir(m1 + 1, md0, # links(m0, dirs0), m2 + 1, entries2), dirs0[m] = mkdir(m8, md5, # links(m, dirs0), m11, entries), dirs0[entries[str]] = mkdir(m13, md6, m12 + 1, m15, ∅), dirs0[entries2[str0]] = mkdir(m4, md3, # links(entries2[str0], dirs0), m7, entries1), files[ino] = mkfile(m14, md8, m16, content), dirs0[ino] = mkdir(# entries4, md9, # links(ino, dirs0), # subdirs(ino, dirs0), entries4), afs-cons(dirs0, files), ROOT_INO ∈ dirs0, entries[str] ∈ dirs0, m0 ∈ dirs0, str0 ∈ entries2, entries2[str0] ∈ dirs0, str ∈ entries, m ∈ dirs0, name ∈ entries5, ino ∈ dirs0, # links(ino, dirs0) ≤ 1, name ∈ entries4 → entries4[name] ∈ dirs0 ∨ entries4[name] ∈ files, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries4 = ∅, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries[str] = ROOT_INO, entries[str] = m, entries[str] = m0, ino = entries[str], m = m0, entries5[name] ∈ files, entries5[name] ∈ dirs0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

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