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Proof step 10 from proof dir-empty (previous step)

dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), tree[p].dir?, tree[p + str '].dir?, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), entries[str] ∈ dirs, str ∈ entries, m ∈ dirs ⊦ entries0 = ∅ ↔ tree[p + str '].entries = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ p ∈ tree → tree[p + q] = tree[p][q]
Proof step 12:
⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files))
Proof step 13:
dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), tree[p][str '].dir?, tree[p].dir?, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), m ∈ dirs, str ∈ entries, entries[str] ∈ dirs, p ∈ tree ⊦ entries0 = ∅ ↔ tree[p][str '].entries = ∅

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));% ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s) Used terms: str ', p, tree