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Proof step 90 from proof abs-paths-eq (previous step)

((m => mkdir(# entries, tree[p3].meta, m1, m2, entries) * abs(tree[p3].entries, entries) * abs|(tree, ino0, p3, m)))(dirs), tree[p3].dir?, tree[p3 + str ' + p0].dir?, ispath(p0, entries[str], m, dirs, files), ispath(p3, ino0, m, dirs, files), str ∈ tree[p3].entries, p3 ∈ tree, ino0 ∈ dirs, p0 ∈ tree[p3][str '], str ' ∈ tree[p3], str ∈ entries, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 91:
tree[p3] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * abs(sdt, entries) * abs|(tree, ino0, p3, m)))(dirs), tree[p3 + str ' + p0].dir?, tree[p3].dir?, ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], m, dirs, files), str ∈ entries, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], ino0 ∈ dirs, p3 ∈ tree, str ∈ sdt, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

Rule argument

spec: directory-tree name: dnode-elim seq: tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt) subst: [tree, md, sdt] → [tree[p3], md, sdt]

Additional information

Used simplifier rules: tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt); (elim) Used terms: tree[p3], md, sdt