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Proof step 824 from proof inv-renam (previous step)

dirs[ino] = mkdir(m1, md1, # links(ino, dirs), m5, entries1), dirs[ino1] = mkdir(m0, md0, # links(ino1, dirs), m3, entries0), dirs[entries0[str]] = mkdir(m, md, # links(entries0[str], dirs), m2, entries), dent = negdentry(str0), err = ESUCCESS, p_ino = ino, dom of < ino2, afs-cons(dirs, files), ispath(q0, ROOT_INO, ino, dirs, files), ispath(q, ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pw(user, md), pw(user, md0), px(user, md0), px(user, md1), px-parent(user, q0, ROOT_INO, dirs, files), px-parent(user, q, ROOT_INO, dirs, files), str ∈ entries0, entries0[str] ∈ dirs, ino1 ∈ dirs, ino ∈ dirs, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ entries0[str] = ROOT_INO, pre-lookup(p_ino, dent, dirs, files), q + str ' ⊑ q0 + str0 '

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (s)% ⊦ ispath(ε, n, m, dirs, files) ↔ valid-ino(n, dirs, files) ∧ m = n; (s)% ⊦ px-parent(user, ε, ino, dirs, files); (s)% ⊦ afs-cons(dirs, files) → ¬ ROOT_INO ∈ files; (s)% ⊦ afs-cons(dirs, files) → ROOT_INO ∈ dirs; (ss)%⊦ ispath(p, ino, n, dirs, files) → (# links(n, dirs) = 0 ↔ p = ε ∧ links(ino, dirs) = ∅); (ws)% ⊦ mkdir(m, md, m0, m1, entries).nlink = m0; (s)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)%⊦ afs-cons(dirs, files) → ( links(ino, dirs) = ∅ ↔ (ino ∈ dirs → dirs[ino].nlink = 0) ∧ (ino ∈ files → files[ino].nlink = 0)); (ws)%⊦ valid-parent-ino(ino, dirs, files) ↔ ino ∈ dirs ∧ (links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO); (s)%⊦ pre-lookup(p_ino, dent, dirs, files) ↔ valid-parent-ino(p_ino, dirs, files); (s)