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Proof step 38 from proof abs-acyclic (previous step)

tree[p + str].dir? ↔ entries[str] ∈ dirs, entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt ⊦ sdt[str].dir?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), tree[p + str '].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt, ROOT_INO ∈ dirs ⊦ sdt[str].dir?

Rule argument

Additional information

Used simplifier rules: ⊦ afs-cons(dirs, files) → ROOT_INO ∈ dirs; (ss)%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]%⊦ ¬ ino ∈ dirs → (((abs(tree, ino) * D))(dirs) ↔ tree = fnode(⌜ino⌝ ) ∧ D(dirs)); (ws)% ⊦ p + str = p + str '; (s)