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Proof step 9 from proof abs-delorphans (previous step)

links(ino, dirs) = ∅, abs(tree, ROOT_INO)(dirs), ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # orphans(dirs, files) = 0, ino = ROOT_INO, orphans(dirs, files) = ∅

User applied rule apply lemma

And obtained the following new goals

Proof step 10:
⊦ abs(tree, ino0)(dirs) ∧ ino1 ∈ dirs → (∃ p. ispath(p, ino0, ino1, dirs, files))
Proof step 11:
links(ino, dirs) = ∅, abs(tree, ROOT_INO)(dirs), ino ∈ dirs, abs(tree, ROOT_INO)(dirs) ∧ ino ∈ dirs → (∃ p. ispath(p, ROOT_INO, ino, dirs, files)), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # orphans(dirs, files) = 0, ino = ROOT_INO, orphans(dirs, files) = ∅

Rule argument

spec: <toplevel> name: abs-has-rootpath seq: ⊦ abs(tree, ino0)(dirs) ∧ ino1 ∈ dirs → (∃ p. ispath(p, ino0, ino1, dirs, files)) subst: [tree, ino0, dirs, ino1, files] → [tree, ROOT_INO, dirs, ino, files]

Additional information

Used terms: tree, ROOT_INO, dirs, ino, files