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Proof step 31 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) * (abs(sdt -- str, entries -- str) * _∃(λ ad. entries[str] => ad * abs(sdt0, ad.entries) ∧ ⌜ad.meta = md1⌝ ∧ ⌜ad.size = # ad.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 ∈ entries, p ∈ tree, entries[str] ∈ dirs, str ∈ sdt ⊦ entries0 = ∅ ↔ sdt0 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
tree[p] = dnode(md, sdt), sdt[str] = dnode(md0, sdt0), (( entries[str] => mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0) * abs(sdt0, entries0) * m => mkdir(# entries, md, # links(m, dirs), n1, entries) * 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, str ∈ sdt ⊦ entries0 = ∅ ↔ sdt0 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).meta = md; (s)% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ mkdir(m, md, m0, m1, entries).size = m; (s)% ⊦ ((ino => ad * D))(dirs) → ino ∈ dirs; (s)% ⊦ ((ino => ad * D))(dirs) → (dirs[ino] = ad0 ↔ ad = ad0); (s)% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (s)%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]%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]% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)% ⊦ ((_∃ adD * D))(dirs) ↔ (∃ ad. ((adD(ad) * D))(dirs)); (s)