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Proof step 18 from proof dir-empty (previous step)

dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), tree[p][str '].dir?, tree[p].dir?, (_∃(λ ino1. abs|(tree, ino, p, ino1) * abs(tree[p], ino1)))(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), m ∈ dirs, str ∈ entries, entries[str] ∈ dirs, p ∈ tree ⊦ entries0 = ∅ ↔ tree[p][str '].entries = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
dirs[entries[str]] = mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0), dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), ((abs|(tree, ino, p, m) * abs(tree[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), m ∈ dirs, str ∈ entries, entries[str] ∈ dirs, p ∈ tree ⊦ entries0 = ∅ ↔ tree[p][str '].entries = ∅

Rule argument

Additional information

Used simplifier rules: absc-path: ⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1; (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]% ⊦ (_∃ nD)(dirs) ↔ (∃ ino. nD(ino)(dirs)); (s)