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Proof step 28 from proof dir-empty (previous step)

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 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
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 -- str, entries -- str) * abs(dnode(md1, sdt0), 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 ∈ sdt, entries[str] ∈ dirs, p ∈ tree, str ∈ entries ⊦ entries0 = ∅ ↔ sdt0 = ∅

Rule argument

Additional information

Used simplifier rules: and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)