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Proof step 25 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), sdt[str].dir?, 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 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
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 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt).dir?; (s)