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Proof step 26 from proof dir-empty (previous step)

tree[p] = dnode(md, sdt), dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), sdt[str] = dnode(md1, sdt0), ((m => mkdir(# entries, md, # links(m, dirs), n1, entries) * abs(sdt, entries) * abs|(tree, ino, p, m)))(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), str ∈ sdt, entries[str] ∈ dirs, p ∈ tree, str ∈ entries ⊦ entries0 = ∅ ↔ sdt0 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ str ∈ sdt → abs(sdt, sdi) = (⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str]))
Proof step 28:
sdt[str] = dnode(md1, sdt0), dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), tree[p] = dnode(md, sdt), (( m => mkdir(# entries, md, # links(m, dirs), n1, entries) * (⌜str ∈ entries⌝ ∧ abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino, p, m)) )(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), str ∈ entries, p ∈ tree, entries[str] ∈ dirs, str ∈ sdt ⊦ entries0 = ∅ ↔ sdt0 = ∅

Rule argument

spec: vfs-abs name: entries-recdel seq: ⊦ str ∈ sdt → abs(sdt, sdi) = ( ⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str])) subst: [str, sdi, sdt] → [str, entries, sdt] flags: <all paths>

Additional information

Used terms: str, entries, sdt