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Proof step 46 from proof abs-acyclic (previous step)

entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ( ( (ROOT_INO => mkdir(# entries0, sdt[str].meta, m0, m1, entries0) * abs(sdt[str].entries, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), sdt[str].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, str ∈ sdt ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 47:
tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, sdt[str] = dnode(md, sdt0), (( ( ( (ROOT_INO => mkdir(# entries0, md, m0, m1, entries0) * abs(sdt0, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), sdt[str].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 ∈ sdt, str ∈ entries ⊦

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] → [sdt[str], md0, sdt0]

Additional information

Used simplifier rules: tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt); (elim) Used terms: sdt[str], md0, sdt0