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Proof step 122 from proof afs-cons-rename-file-negdentry (previous step)

(m0 = m ⊃ mkdir(m2, md2, # links(m, dirs0), m5, entries1 -- str);mkdir(m7, md5, # links(m0, dirs0), m9, entries2)) = mkdir(m1, md4, m4, m6, entries0), dirs0[m] = mkdir(m2 + 1, md2, # links(m, dirs0), m5, entries1), files[entries1[str]] = mkfile(m3, md, # links(entries1[str], dirs0), content0), dirs0[m0] = mkdir(m7, md5, # links(m0, dirs0), m9, entries2), files[ino] = mkfile(m8, md3, # links(ino, dirs0), content), dirs0[ino] = mkdir(m11, md6, m12, m13, entries), afs-cons(dirs0, files), ino ∈ files, ROOT_INO ∈ dirs0, m0 ∈ dirs0, str ∈ entries1, entries1[str] ∈ files, m ∈ dirs0, m0 = m → md1 = md0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ # links(ino, dirs0) = # links (ino, dirs0 [m, (m = m0 ⊃ mkdir(m1 + 1, md4, m4, m6, entries0[str0, entries1[str]]);mkdir(m2, md2, # links(m, dirs0), m5, entries1 -- str)).meta:= md0 ] [m0, (m0 = m ⊃ mkdir(m1 + 1, md0, m4, m6, entries0[str0, entries1[str]]);mkdir(m1 + 1, md4, m4, m6, entries0[str0, entries1[str]])).meta:= md1 ]), ino ∈ dirs0, entries1[str] ∈ dirs0, str0 ∈ entries2

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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