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Proof step 543 from proof inv-renam (previous step)

dirs[ino0] = mkdir(m4, md2, # links(ino0, dirs), m7, entries2), dirs[ino1] = mkdir(m0, md0, # links(ino1, dirs), m3, entries0), dirs[entries0[str]] = mkdir(m, md, # links(entries0[str], dirs), m2, entries), err = ESUCCESS, ino = mkdentry(str0, entries2[str0]).ino, inode = p_inode, p_inode = mkinode(entries0[str], md, true, # links(entries0[str], dirs), m2, m), mkdentry(str0, entries2[str0]).dentry?, dom of < ino2, afs-cons(dirs, files), ispath(q0, ROOT_INO, ino0, dirs, files), ispath(q, ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pw(user, md2), pw(user, md), pw(user, md0), px(user, md0), px(user, md2), px-parent(user, q0, ROOT_INO, dirs, files), px-parent(user, q, ROOT_INO, dirs, files), str0 ∈ entries2, entries0[str] ∈ dirs, ino1 ∈ dirs, ino0 ∈ dirs, str ∈ entries0, links(ino0, dirs) ≠ ∅ ∨ ino0 = ROOT_INO, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ entries0[str] = ROOT_INO, pre-iget(ino, dirs, files), q + str ' ⊑ q0 + str0 '

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)% ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (s)% ⊦ pre-iget(ino, dirs, files) ↔ valid-ino(ino, dirs, files); (s)% ⊦ mkdentry(str, m).ino = m; (s)% ⊦ mkdentry(str, m).dentry?; (s)