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Proof step 89 from proof afs-cons-rename-dir-negdentry (previous step)

(m0 = m ⊃ mkdir(m1, md, # links(m, dirs0), m2, entries2 -- str);mkdir(m8, md5, # links(m0, dirs0), m10, entries3)) = mkdir(m4, md3, m6, m7, entries0), (ino = m0 ⊃ (m0 = m ⊃ mkdir(m4 + 1, md0, m6, m7 + 1, entries0[str0, entries2[str]]);mkdir(m4 + 1, md3, m6, m7 + 1, entries0[str0, entries2[str]])).meta:= md1 ;(ino = m ⊃ mkdir(m1, md0, # links(m, dirs0), m2, entries2 -- str) ;mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries) ) ) = mkdir(m15, md7, m16, m17, entries4), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), dirs0[m0] = mkdir(m8, md5, # links(m0, dirs0), m10, entries3), dirs0[m] = mkdir(m1 + 1, md, # links(m, dirs0), m2 + 1, entries2), files[ino] = mkfile(m9, md2, m11, content), dirs0[ino] = mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries), links (ino, dirs0 [m, (m = m0 ⊃ mkdir(m4 + 1, md3, m6, m7 + 1, entries0[str0, entries2[str]]);mkdir(m1, md, # links(m, dirs0), m2, entries2 -- str)).meta:= md0 ] [m0, (m0 = m ⊃ mkdir(m4 + 1, md0, m6, m7 + 1, entries0[str0, entries2[str]]);mkdir(m4 + 1, md3, m6, m7 + 1, entries0[str0, entries2[str]])).meta:= md1 ]) = ∅, afs-cons(dirs0, files), ino ∈ dirs0, ROOT_INO ∈ dirs0, m0 ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, m ∈ dirs0, # links(ino, dirs0) ≤ 1, m0 = m → md1 = md0, name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files, links(ino, dirs0) = ∅ → entries = ∅, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries4 = ∅, ino = ROOT_INO, str0 ∈ entries3

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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