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Proof step 56 from proof abs-nlinks-h (previous step)

dirs[m] = mkdir(m0, md, m1, m2, entries), abs(tree, ino0)(dirs), ispath(p, ino0, m, dirs, files), str ∈ entries, m ∈ dirs, ino0 ∈ dirs, entries[str] ∈ files, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # links(entries[str], dirs) = 0, ⌜entries[str]⌝ ∈ tree.fids

User applied rule apply lemma

And obtained the following new goals

Proof step 57:
∀ 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 58:
ispath(p, ino0, dirs, files) ↔ p ∈ tree, dirs[m] = mkdir(m0, md, m1, m2, entries), abs(tree, ino0)(dirs), ispath(p, ino0, m, dirs, files), entries[str] ∈ files, ino0 ∈ dirs, m ∈ dirs, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(entries[str], dirs) = 0, ⌜entries[str]⌝ ∈ tree.fids

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, tree, ino0, emp, p]

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, p