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Proof step 153 from proof afs-cons-rename-file-dentry (previous step)

# links(entries[str], dirs0) = m11 + 1, (ino = entries[str] ⊃ mkfile(m2, md, m11, content1);mkfile(m12, md7, # links(ino, dirs0), content0)) = mkfile(m14, md8, m15, content2), dirs0[ino] = mkdir(m16, md9, m17, m18, entries2), dirs0[m0] = mkdir(m3 + 1, md4, # links(m0, dirs0), m7, entries1), files0[entries1[str0]] = mkfile(m1, md3, # links(entries1[str0], dirs0), content), files0[entries[str]] = mkfile(m5, md0, m11 + 1, content1), dirs0[m] = mkdir(m6, md6, # links(m, dirs0), m10, entries), files0[ino] = mkfile(m12, md7, # links(ino, dirs0), content0), afs-cons(dirs0, files0), ROOT_INO ∈ dirs0, entries[str] ∈ files0, m0 ∈ dirs0, str ∈ entries, entries1[str0] ∈ files0, str0 ∈ entries1, ino ∈ files0, m ∈ dirs0, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ m0 = m, m15 = # (entries1[str0] = ino ⊃ (links(entries1[str0], dirs0) -- mklink(m0, str0)) ++ mklink(m, str) ;(entries[str] = ino ⊃ links(entries[str], dirs0) -- mklink(m, str);links(ino, dirs0)) ), entries[str] ∈ dirs0, entries1[str0] ∈ dirs0, ino ∈ dirs0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

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