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Proof step 18 from proof abs-nlinks-h (previous step)

tree[p0 + str '] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, abs(tree, ino0)(dirs), ispath(p0 + str ', ino0, m, dirs, files), ino1 ∈ files, p0 ∈ tree, str ' ∈ tree[p0], ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ ispath(p + q, n, m, dirs, files) ↔ (∃ m0. ispath(p, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files))
Proof step 20:
tree[p0 + str '] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, abs(tree, ino0)(dirs), ino0 ∈ dirs, str ' ∈ tree[p0], p0 ∈ tree, ino1 ∈ files, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∃ m0. ispath(p0, ino0, m0, dirs, files) ∧ ispath(str ', m0, m, dirs, files) ⊦

Rule argument

spec: afs-base name: path-split seq: ⊦ ispath(p + q, n, m, dirs, files) ↔ (∃ m0. ispath(p, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files)) subst: [files, dirs, m, n, q, p] → [files, dirs, m, ino0, str ', p0] flags: <all paths>

Additional information

Used terms: files, dirs, m, ino0, str ', p0