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Proof step 7 from proof abs-nlinks-h (previous step)

tree[p] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, abs(tree, ino0)(dirs), p ∈ tree, ino0 ∈ dirs, ino1 ∈ files, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 8:
∀ 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 9:
ispath(p, ino0, dirs, files) ↔ p ∈ tree, tree[p] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, abs(tree, ino0)(dirs), ino1 ∈ files, ino0 ∈ dirs, p ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

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