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Proof step 237 from proof afs-cons-rename-dir-negdentry (previous step)

# links(m0, dirs0) = dirs0[m0].nlink, # links(entries2[str], dirs0) = m2, dirs0[m] = mkdir(0, md, # links(m, dirs0), m3, entries2), dirs0[m0] = mkdir(dirs0[m0].size, md2, dirs0[m0].nlink, dirs0[m0].nsubdirs, entries), dirs0[entries2[str]] = mkdir(m1, md3, m2, m4, entries0), dirs0[entries2[str]] = mkdir(m1, md4, m2, m4, entries1), afs-cons(dirs0, files), str ∈ entries2, entries2[str] ∈ dirs0, m ∈ dirs0, m0 ∈ dirs0, m0 = m → md1 = md0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ str0 ∈ entries, str0 ∈ dirs0[m0].entries

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 238:
# links(m0, dirs0) = dirs0[m0].nlink, dirs0[entries2[str]] = mkdir(m1, md4, # links(entries2[str], dirs0), m4, entries1), dirs0[m0] = mkdir(dirs0[m0].size, md2, dirs0[m0].nlink, dirs0[m0].nsubdirs, entries), dirs0[m] = mkdir(0, md, # links(m, dirs0), m3, entries2), afs-cons(dirs0, files), m0 ∈ dirs0, m ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, m0 = m → md1 = md0, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ str0 ∈ dirs0[m0].entries, str0 ∈ entries

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)