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Proof step 160 from proof afs-cons-rename-dir-negdentry (previous step)

# entries2 = m4 + 1, # subdirs(m0, dirs0) = m7 + 1, files[m0] = mkfile(m9, md2, m11, content), dirs0[m0] = mkdir(m4 + 1, md, # links(m0, dirs0), m7 + 1, entries2), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), afs-cons(dirs0, files), m0 ∈ dirs0, ROOT_INO ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, # links(m0, dirs0) ≤ 1, name ∈ entries2 → entries2[name] ∈ dirs0 ∨ entries2[name] ∈ files, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries2 = ∅, m7 = #(subdirs(m0, dirs0) -- str), str0 ∈ entries2, m0 ≠ ROOT_INO ∧ links(m0, dirs0) = ∅

Heuristic use patterns applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-union, inst: dir-entries name: del-size seq: ⊦ str ∈ strings → #(strings -- str) = # strings - 1 subst: [str, strings] → [str, subdirs(m0, dirs0)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ str ∈ strings → strings -- str = strings; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s) Used terms: str, subdirs(m0, dirs0)