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Proof step 9 from proof abs-paths-eq (previous step)

abs(tree, ino0)(dirs), ispath(p0, ino0, ino1, dirs, files), ispath(p1, ino0, ino1, dirs, files), ino1 ∈ dirs, p0 ∈ tree, p1 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p0 = p1

User applied rule apply lemma

And obtained the following new goals

Proof step 10:
∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ((abs(tree, ino) * D))(dirs), ispath(p, ino, ino0, dirs, files) ⊦ tree[p].dir? ↔ ino0 ∈ dirs
Proof step 11:
tree[p1].dir? ↔ ino1 ∈ dirs, abs(tree, ino0)(dirs), ispath(p1, ino0, ino1, dirs, files), ispath(p0, ino0, ino1, dirs, files), p1 ∈ tree, p0 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p0 = p1

Rule argument

spec: <toplevel> name: abs-dir-h seq: ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ((abs(tree, ino) * D))(dirs), ispath(p, ino, ino0, dirs, files) ⊦ tree[p].dir? ↔ ino0 ∈ dirs subst: [files, dirs, tree, ino, D, p, ino0] → [files, dirs, tree, ino0, emp, p1, ino1]

Additional information

Used simplifier rules: ⊦ emp * D = D; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap] Used terms: files, dirs, tree, ino0, emp, p1, ino1