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Proof step 311 from proof afs-cons-rename-dir-dentry (previous step)

# links(entries[str], dirs0) = dirs0[entries[str]].nlink, # links(m, dirs0) = dirs0[m].nlink, # links(entries2[str0], dirs0) = m2, dirs0[entries[str]].entries = ∅, dirs0[m0] = mkdir(0, md, # links(m0, dirs0), m3, entries2), dirs0[m] = mkdir(dirs0[m].size, md4, dirs0[m].nlink, dirs0[m].nsubdirs, entries), dirs0[entries[str]] = mkdir(dirs0[entries[str]].size, md3, dirs0[entries[str]].nlink, dirs0[entries[str]].nsubdirs, ∅), dirs0[entries2[str0]] = mkdir(m1, md5, m2, m4, entries0), entries[str] = dirs0[m].entries[str], dirs0[entries2[str0]] = mkdir(m1, md0, m2, m4, entries1), afs-cons(dirs0, files), m0 ∈ dirs0, m ∈ dirs0, entries2[str0] ∈ dirs0, entries[str] ∈ dirs0, str0 ∈ entries2, str ∈ entries, str ∈ dirs0[m].entries, m = m0 → md1 = md2, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries[str] = ROOT_INO, m0 = m ∧ mkdentry(str, entries[str]) = mkdentry(str0, entries2[str0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 312:
# links(m, dirs0) = dirs0[m].nlink, # links(entries[str], dirs0) = dirs0[entries[str]].nlink, dirs0[entries[str]].entries = ∅, entries[str] = dirs0[m].entries[str], dirs0[entries2[str0]] = mkdir(m1, md0, # links(entries2[str0], dirs0), m4, entries1), dirs0[entries[str]] = mkdir(dirs0[entries[str]].size, md3, dirs0[entries[str]].nlink, dirs0[entries[str]].nsubdirs, ∅), dirs0[m] = mkdir(dirs0[m].size, md4, dirs0[m].nlink, dirs0[m].nsubdirs, entries), dirs0[m0] = mkdir(0, md, # links(m0, dirs0), m3, entries2), afs-cons(dirs0, files), str ∈ dirs0[m].entries, str ∈ entries, str0 ∈ entries2, entries[str] ∈ dirs0, entries2[str0] ∈ dirs0, m ∈ dirs0, m0 ∈ dirs0, m = m0 → md1 = md2, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries[str] = ROOT_INO, m0 = m ∧ mkdentry(str, entries[str]) = mkdentry(str0, entries2[str0])

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)