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Proof step 20 from proof abs-unshared-h (previous step)

(((abs|(tree1, ino1, p1, n1) * abs(tree1[p1], n1)) * abs|(tree0, ino0, p0, n1) * abs(tree0[p0], n1) * D))(dirs), ispath(p0, ino0, n1, dirs, files), ispath(p1, ino1, n1, dirs, files), p0 ∈ tree0, n1 ∈ dirs, p1 ∈ tree1, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ tree0[p0].dir?

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid))
Proof step 22:
(((abs|(tree1, ino1, p1, n1) * abs(tree1[p1], n1)) * abs|(tree0, ino0, p0, n1) * abs(tree0[p0], n1) * D))(dirs), ispath(p1, ino1, n1, dirs, files), ispath(p0, ino0, n1, dirs, files), p1 ∈ tree1, n1 ∈ dirs, p0 ∈ tree0, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ∀ fid. tree0[p0] ≠ fnode(fid)

Rule argument

spec: directory-tree name: fnode-elim-strong seq: ⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid)) subst: [tree] → [tree0[p0]] flags: <all paths>

Additional information

Used terms: tree0[p0]