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Proof step 4 from proof path-in (previous step)

abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ tree[p].dir? ↔ m ∈ dirs

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
∀ 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 6:
⊦ ⌜n⌝ ∈ tree.fids ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) → n ∈ files
Proof step 7:
ispath(p, ino, dirs, files) ↔ p ∈ tree, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ tree[p].dir? ↔ m ∈ dirs

Rule argument

spec: vfs-abs 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, tree, ino, emp, p]

Additional information

Used simplifier rules: fid-file: ⊦ ⌜n⌝ ∈ tree.fids ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) → n ∈ files;% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)% ⊦ 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, ino, emp, p