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Proof step 3 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, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ((abs(tree, ino) * D))(dirs) ⊦ ispath(p, ino, dirs, files) ↔ p ∈ tree
Proof step 5:
ispath(p0, ino0, dirs, files) ↔ p0 ∈ tree0, ((abs(tree0, ino0) * abs(tree1, ino1) * D))(dirs), ispath(p1, ino1, n1, dirs, files), ispath(p0, ino0, n1, dirs, files), n1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

Rule argument

spec: <toplevel> name: abs-path-h seq: ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ((abs(tree, ino) * D))(dirs) ⊦ ispath(p, ino, dirs, files) ↔ p ∈ tree subst: [files, dirs, tree, ino, D, p] → [files, dirs, tree0, ino0, abs(tree1, ino1) * D, p0]

Additional information

Used terms: files, dirs, tree0, ino0, abs(tree1, ino1) * D, p0