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Proof step 175 from proof afs-cons-rename-file-dentry (previous step)

# links(entries2[str], dirs0) = m11 + 1, files0[m] = mkfile(m12, md7, m13, content0), dirs0[m] = mkdir(# entries2, md9, # links(m, dirs0), # subdirs(m, dirs0), entries2), files0[entries2[str]] = mkfile(m5, md0, m11 + 1, content1), files0[entries1[str0]] = mkfile(m1, md3, # links(entries1[str0], dirs0), content), dirs0[m0] = mkdir(m3 + 1, md4, # links(m0, dirs0), m7, entries1), afs-cons(dirs0, files0), name ∈ entries2, m ∈ dirs0, name ∈ entries2[str, entries1[str0]], str0 ∈ entries1, entries1[str0] ∈ files0, str ∈ entries2, m0 ∈ dirs0, entries2[str] ∈ files0, ROOT_INO ∈ dirs0, # links(m, dirs0) ≤ 1, entries2[name] ∈ dirs0 ∨ entries2[name] ∈ files0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries2 = ∅, m = m0, name = str, entries1[str0] ∈ dirs0, entries2[str] ∈ dirs0, entries2[str, entries1[str0]][name] ∈ dirs0, entries2[str, entries1[str0]][name] ∈ files0, m ≠ ROOT_INO ∧ links(m, dirs0) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 176:
# links(entries2[str], dirs0) = m11 + 1, dirs0[m0] = mkdir(m3 + 1, md4, # links(m0, dirs0), m7, entries1), files0[entries1[str0]] = mkfile(m1, md3, # links(entries1[str0], dirs0), content), files0[entries2[str]] = mkfile(m5, md0, m11 + 1, content1), dirs0[m] = mkdir(# entries2, md9, # links(m, dirs0), # subdirs(m, dirs0), entries2), files0[m] = mkfile(m12, md7, m13, content0), afs-cons(dirs0, files0), m ∈ dirs0, ROOT_INO ∈ dirs0, entries2[str] ∈ files0, m0 ∈ dirs0, str ∈ entries2, entries1[str0] ∈ files0, str0 ∈ entries1, name ∈ entries2, entries2[name] ∈ dirs0, # links(m, dirs0) ≤ 1, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries2 = ∅, m = m0, entries2[str, entries1[str0]][name] ∈ dirs0, entries2[str] ∈ dirs0, entries1[str0] ∈ dirs0, m ≠ ROOT_INO ∧ links(m, dirs0) = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)% ⊦ str ≠ str0 → entries[str0, ino][str] = entries[str]; (s)% ⊦ str ≠ str0 → (str ∈ entries[str0, ino] ↔ str ∈ entries); (s)