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Proof step 128 from proof afs-cons-rename-dir-dentry (previous step)

# links(entries2[str], dirs0) = m12 + 1, (ino = entries2[str] ⊃ mkdir(m13, md, m12, m15, ∅) ;(ino = m ⊃ mkdir(m8, md1, # links(m, dirs0), m11, (entries2 -- str0)[str, entries2[str0]]) ;mkdir(# entries4, md9, # links(ino, dirs0), # subdirs(ino, dirs0), entries4) ) ) = mkdir(m20, md10, m21, m22, entries5), dirs0[m] = mkdir(m8 + 1, md0, # links(m, dirs0), m11 + 1, entries2), dirs0[entries2[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, entries2[str] ∈ dirs0, str0 ∈ entries2, entries2[str0] ∈ dirs0, str ∈ entries2, m ∈ dirs0, ino ∈ dirs0, # links(ino, dirs0) ≤ 1, name ∈ entries4 → entries4[name] ∈ dirs0 ∨ entries4[name] ∈ files, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries4 = ∅, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries2[str] = ROOT_INO, entries2[str] = m, m21 = # (entries2[str0] = ino ⊃ (links(entries2[str0], dirs0) -- mklink(m, str0)) ++ mklink(m, str) ;(entries2[str] = ino ⊃ links(entries2[str], dirs0) -- mklink(m, str);links(ino, dirs0)) ), mkdentry(str, entries2[str]) = mkdentry(str0, entries2[str0])

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

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