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Proof step 17 from proof abs-acyclic (previous step)

entries[str] = ROOT_INO, dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m0, entries), (((ino0 => mkdir(# entries0, tree[p].meta, m1, m2, entries0) * abs(tree[p].entries, entries0)) * abs|(tree, ROOT_INO, p, ino0)))(dirs), tree[p].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs, str ∈ entries ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m0, entries), entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (((ino0 => mkdir(# entries0, md0, m1, m2, entries0) * abs(sdt, entries0)) * abs|(tree, ROOT_INO, p, ino0)))(dirs), tree[p].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, ino ∈ dirs ⊦

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[p], md0, sdt]

Additional information

Used simplifier rules: tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt); (elim) Used terms: tree[p], md0, sdt