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Proof step 203 from proof afs-cons-rename-dir-negdentry (previous step)

# links(m, dirs0) = dirs0[m].nlink, # links(m0, dirs0) = m4, dirs0[m].nsubdirs = 0, dirs0[m] = mkdir(dirs0[m].size, md4, dirs0[m].nlink, 0, entries1), dirs0[m0] = mkdir(m2, md2, m4, m5, entries), dirs0[entries1[str]] = mkdir(m1, md, # links(entries1[str], dirs0), m3, entries2), entries1[str] = dirs0[m].entries[str], dirs0[m0] = mkdir(m2, md3, m4, m5, entries0), afs-cons(dirs0, files), entries1[str] ∈ dirs0, str ∈ entries1, m ∈ dirs0, m0 ∈ dirs0, str ∈ dirs0[m].entries, 1 ≤ dirs0[m].size, m0 = m → md1 = md0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ dirs0[m].size = 0, str0 ∈ entries, str0 ∈ entries0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 204:
# links(m, dirs0) = dirs0[m].nlink, dirs0[m].nsubdirs = 0, entries1[str] = dirs0[m].entries[str], dirs0[entries1[str]] = mkdir(m1, md, # links(entries1[str], dirs0), m3, entries2), dirs0[m0] = mkdir(m2, md3, # links(m0, dirs0), m5, entries0), dirs0[m] = mkdir(dirs0[m].size, md4, dirs0[m].nlink, 0, entries1), afs-cons(dirs0, files), str ∈ dirs0[m].entries, m0 ∈ dirs0, m ∈ dirs0, str ∈ entries1, entries1[str] ∈ dirs0, 1 ≤ dirs0[m].size, m0 = m → md1 = md0, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ dirs0[m].size = 0, str0 ∈ entries0

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)