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Proof step 209 from proof afs-cons-rename-file-dentry (previous step)

# links(m0, dirs0) = m4, # links(m, dirs0) = dirs0[m].nlink, dirs0[m0] = mkdir(0, md6, m4, m5, entries0), files[entries0[str0]] = mkfile(m1, md0, # links(entries0[str0], dirs0), content), files[entries[str]] = mkfile(m2, md, # links(entries[str], dirs0), content1), dirs0[m] = mkdir(dirs0[m].size, md5, dirs0[m].nlink, dirs0[m].nsubdirs, entries), entries[str] = dirs0[m].entries[str], entries0[str0] = entries1[str0], dirs0[m0] = mkdir(m3, md3, m4, m5, entries1), m3 = 0, afs-cons(dirs0, files), entries0[str0] ∈ files, entries[str] ∈ files, m0 ∈ dirs0, m ∈ dirs0, str0 ∈ entries0, str ∈ entries, str ∈ dirs0[m].entries, str0 ∈ entries1, m = m0 → md1 = md2, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries[str] ∈ dirs0, m0 = m ∧ mkdentry(str, entries[str]) = mkdentry(str0, entries0[str0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 210:
# links(m, dirs0) = dirs0[m].nlink, entries[str] = dirs0[m].entries[str], dirs0[m] = mkdir(dirs0[m].size, md5, dirs0[m].nlink, dirs0[m].nsubdirs, entries), files[entries[str]] = mkfile(m2, md, # links(entries[str], dirs0), content1), files[entries1[str0]] = mkfile(m1, md0, # links(entries1[str0], dirs0), content), dirs0[m0] = mkdir(0, md3, # links(m0, dirs0), m5, entries1), afs-cons(dirs0, files), str ∈ dirs0[m].entries, str ∈ entries, str0 ∈ entries1, m ∈ dirs0, m0 ∈ dirs0, entries[str] ∈ files, entries1[str0] ∈ files, m = m0 → md1 = md2, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries[str] ∈ dirs0, m0 = m ∧ mkdentry(str, entries[str]) = mkdentry(str0, entries1[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)