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Proof step 103 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), (( (abs|(sdt[str], entries[str], p0, m) * abs(sdt[str][p0], m)) * m => mkdir(# entries, md, m1, m2, entries) * abs(sdt -- str, entries -- str) * abs|(tree, ino0, p3, m)) )(dirs), tree[p3 + str ' + p0].dir?, ispath(p0, entries[str], m, dirs, files), ispath(p3, ino0, m, dirs, files), str ∈ entries, p3 ∈ tree, ino0 ∈ dirs, str ∈ sdt, p0 ∈ sdt[str], ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ sdt[str][p0].dir?

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: directory-tree name: split-at seq: ⊦ p ∈ tree → tree[p + q] = tree[p][q] subst: [q, p, tree] → [str ' + p0, p3, tree] flags: <all paths>

Additional information

Used simplifier rules: absc-emp: ⊦ abs|(tree, ino, ε, ino0) = (emp ∧ ⌜ino = ino0⌝ );%absc-path: ⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1;% ⊦ tree[ε] = tree; (s)% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)% ⊦ emp * D = D; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (s)% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ p + ε = p; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]%str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s)% ⊦ ε ∈ tree; (s)%path-in-dirs: p ≠ ε ⊦ ispath(p, n, m, dirs, files) → n ∈ dirs; (forward) from specification afs-base% ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]; (s) Used terms: str ' + p0, p3, tree