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Proof step 9 from proof abs-unshared-h (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1))
Proof step 11:
((_∃(λ ino1. abs|(tree0, ino0, p0, ino1) * abs(tree0[p0], ino1)) * abs(tree1, ino1) * 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 ⊦

Rule argument

spec: <toplevel> name: path-unfold seq: ⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)) subst: [p, ino0, tree] → [p0, ino0, tree0] flags: <all paths>

Additional information

Used terms: p0, ino0, tree0