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Proof step 135 from proof inv-link (previous step)

dirs[ino1] = mkdir(m1, md1, # links(ino1, dirs), m4, entries), files[entries0[str]] = mkfile(m, md, # links(entries0[str], dirs), content), dirs[ino0] = mkdir(m0, md0, # links(ino0, dirs), m3, entries0), dent = negdentry(p.last), err = ESUCCESS, err0 = EEXISTS, err1 = ESUCCESS, p_ino = ino1, afs-cons(dirs, files), ispath(q, ROOT_INO, ino0, dirs, files), ispath(parent(p), ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pre-lookup(p_ino, dent, dirs, files), pw(user, md1), px(user, md0), px(user, md1), px-parent(user, q, ROOT_INO, dirs, files), px-parent(user, parent(p), ROOT_INO, dirs, files), str ∈ entries0, entries0[str] ∈ files, ino0 ∈ dirs, ino1 ∈ dirs, dent.name ∈ dirs[p_ino].entries, links(ino0, dirs) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ ⟪skip⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files)), p = ε, entries0[str] ∈ dirs

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ negdentry(str).name = str; (s)%⊦ pre-lookup(p_ino, dent, dirs, files) ↔ valid-parent-ino(p_ino, dirs, files); (s)%⊦ valid-parent-ino(ino, dirs, files) ↔ ino ∈ dirs ∧ (links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO); (s)