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Proof step 9 from proof pr (previous step)

tree[p].dir? ↔ m ∈ dirs, meta(p, tree, abs-files(files)) = meta(ε, m, dirs, files), 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) ⊦ pr(user, p, tree, abs-files(files)) ↔ pr(user, meta(ε, m, dirs, files))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: path-permission, inst: posix-permission name: pr seq: ⊦ pr(user, p, tree, fs) ↔ pr(user, meta(p, tree, fs)) ∧ px-parent(user, p, tree, fs) subst: [fs, tree, p, user] → [abs-files(files), tree, p, user] flags: <all paths>

Additional information

Used simplifier rules: px-parent: ⊦ ispath(p, ino, m, dirs, files) ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → ( px-parent(user, p, tree, abs-files(files)) ↔ px-parent(user, p, ino, dirs, files));%path-in-dirs: p ≠ ε ⊦ ispath(p, n, m, dirs, files) → n ∈ dirs; (forward) from specification afs-base Used terms: abs-files(files), tree, p, user