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Proof step 39 from proof abs-acyclic (previous step)

entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), tree[p + str '].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt, ROOT_INO ∈ dirs ⊦ sdt[str].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 ', p, tree] flags: <all paths>

Additional information

Used simplifier rules: path-in-eq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files));%path-in-eq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files));% ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)%str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s) Used terms: str ', p, tree