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Proof step 589 from proof inv-renam (previous step)

isdir1 ↔ true, isdir ↔ true, isdir0 ↔ isdir, 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 = mkdentry(str0, entries1[str0]), err = err0, inode = mkinode(ino, md1, true, # links(ino, dirs), m5, m1), p_inode = mkinode(entries0[str], md, true, # links(entries0[str], dirs), m2, m), 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, str0 ∈ entries1, ⌊err0⌋ , links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ ⟪skip⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files)), entries0[str] = ROOT_INO, q + str ' ⊑ q0 + str0 '

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

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