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Proof step 255 from proof inv-link (previous step)

dirs[ino0] = mkdir(m0, md0, # links(ino0, dirs), m3, entries0), files[entries0[str]] = mkfile(m, md, # links(entries0[str], dirs), content), dent = negdentry(p.last), err = err0, err0 = ESUCCESS, ino = ino1, inode = mkinode(entries0[str], md, false, # links(entries0[str], dirs), 0, m), afs-cons(dirs, files), ispath(parent(p), ROOT_INO, ino1, dirs, files), ispath(q, ROOT_INO, ino0, dirs, files), of-cons(of, dirs, files), px(user, md0), px-parent(user, parent(p), ROOT_INO, dirs, files), px-parent(user, q, ROOT_INO, dirs, files), ino0 ∈ dirs, entries0[str] ∈ files, str ∈ entries0, links(ino0, dirs) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ p = ε, pre-iget(ino, dirs, files), entries0[str] ∈ dirs

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: path-in-end: ⊦ ispath(p, n, m, dirs, files) ∧ ¬ m ∈ files → m ∈ dirs; (forward) from specification afs-base% ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (s)% ⊦ pre-iget(ino, dirs, files) ↔ valid-ino(ino, dirs, files); (s)