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Proof step 75 from proof inv-rmdir (previous step)

# subdirs(m, dirs0) = m14 + 1, # entries0 = m3 + 1, # links(entries0[str], dirs0) = m2 + 1, dirs0[entries0[str]] = mkdir(m5, md1, m2 + 1, m7, ∅), files[m] = mkfile(m6, md, m8, content), dirs0[m] = mkdir(m3 + 1, md3, # links(m, dirs0), m14 + 1, entries0), afs-cons(dirs0, files), entries0[str] ∈ dirs0, ROOT_INO ∈ dirs0, m ∈ dirs0, str ∈ entries0, # links(m, dirs0) ≤ 1, name ∈ entries0 → entries0[name] ∈ dirs0 ∨ entries0[name] ∈ files, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries0[str] = ROOT_INO, entries0[str] = m, entries0 = ∅, m14 = #(subdirs(m, dirs0) -- str), m ≠ ROOT_INO ∧ links(m, 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(m, 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(m, dirs0)