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Proof step 211 from proof afs-cons-rename-file-dentry (previous step)

# links(m, dirs0) = m4, dirs0[m0] = mkdir(0, md3, # links(m0, dirs0), m5, entries1), files[entries1[str0]] = mkfile(m1, md0, # links(entries1[str0], dirs0), content), files[entries[str]] = mkfile(m2, md, # links(entries[str], dirs0), content1), dirs0[m] = mkdir(m3, md5, m4, m6, entries), entries[str] = entries0[str], dirs0[m] = mkdir(m3, md4, m4, m6, entries0), afs-cons(dirs0, files), entries1[str0] ∈ files, entries[str] ∈ files, m0 ∈ dirs0, m ∈ dirs0, str0 ∈ entries1, str ∈ entries, str ∈ entries0, 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, entries1[str0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 212:
dirs0[m] = mkdir(m3, md4, # links(m, dirs0), m6, entries0), files[entries0[str]] = mkfile(m2, md, # links(entries0[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 ∈ entries0, str0 ∈ entries1, m ∈ dirs0, m0 ∈ dirs0, entries0[str] ∈ files, entries1[str0] ∈ files, m = m0 → md1 = md2, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries0[str] ∈ dirs0, m0 = m ∧ mkdentry(str, entries0[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)