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Proof step 23 from proof dir-empty (previous step)

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) * abs(sdt, entries) * abs|(tree, ino, p, m)))(dirs), tree[p][str '].dir?, tree[p].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 ∈ entries, p ∈ tree, entries[str] ∈ dirs, str ∈ sdt ⊦ entries0 = ∅ ↔ tree[p][str '].entries = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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) * 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 ∈ entries, p ∈ tree, entries[str] ∈ dirs, str ∈ sdt ⊦ entries0 = ∅ ↔ sdt[str].entries = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt).dir?; (s)%str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s)%entries-dom1: ⊦ str ∈ sdi ∧ ((abs(sdt, sdi) * D))(dirs) → str ∈ sdt; (forward) from specification vfs-abs%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]