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Proof step 180 from proof afs-cons-rename-dir-negdentry (previous step)

# subdirs(m, dirs0) = m7 + 1, # entries3 = m4 + 1, files[m] = mkfile(m9, md2, m11, content), dirs0[m] = mkdir(m4 + 1, md5, # links(m, dirs0), m7 + 1, entries3), dirs0[entries3[str]] = mkdir(m3, md4, # links(entries3[str], dirs0), m5, entries1), afs-cons(dirs0, files), ROOT_INO ∈ dirs0, entries3[str] ∈ dirs0, str ∈ entries3, m ∈ dirs0, name ∈ entries3, # links(m, dirs0) ≤ 1, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, entries3[name] ∈ dirs0 ∨ entries3[name] ∈ files ⊦ entries3 = ∅, name = str, str0 ∈ entries3, (entries3 -- str)[str0, entries3[str]][name] ∈ dirs0, (entries3 -- str)[str0, entries3[str]][name] ∈ files, m ≠ ROOT_INO ∧ links(m, dirs0) = ∅

Heuristic use patterns applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: store, inst: dir-entries name: At-other seq: ⊦ str ≠ str0 → entries[str0, ino][str] = entries[str] subst: [str, str0, entries, ino] → [name, str0, entries3 -- str, entries3[str]] flags: <all paths>

Additional information

Used simplifier rules: Del-at-entries: ⊦ (entries -- str0)[str] = (str ≠ str0 ⊃ entries[str];(∅)[str]);% ⊦ str ≠ str0 → (entries -- str0)[str] = entries[str]; (s)%⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss) Used terms: name, str0, entries3 -- str, entries3[str]