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Proof step 120 from proof afs-cons-rename-dir-negdentry (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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