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Proof step 200 from proof inv-unlink (previous step)

# links(entries[str], dirs0) = m1 + 1, files0[entries[str]] = mkfile(m0, md, m1 + 1, content), dirs0[ino1] = mkdir(m, md1, # links(ino1, dirs0), m2, entries), dirs = dirs0[ino1, mkdir(m - 1, md1, # links(ino1, dirs0), m2, entries -- str)], files = files0[entries[str], mkfile(m0, md, m1, content)], ino = inode.ino, inode = mkinode(entries[str], md, false, # links(entries[str], dirs0), 0, m0), dom of < ino0, afs-cons(dirs0, files0), ispath(q, ROOT_INO, ino1, dirs0, files0), of-cons(of, dirs0, files0), pw(user, md1), px(user, md1), px-parent(user, q, ROOT_INO, dirs0, files0), entries[str] ∈ files0, str ∈ entries, ino1 ∈ dirs0, 1 ≤ m, links(ino1, dirs0) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ [c_inode := getinode(entries[str], dirs, files); err := ESUCCESS; asynced := ?] (afs-cons(dirs, files) → ⟪vfs_put_inode#(c_inode, of, asynced; err, dirs, files)⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files))), m = 0, ino ∈ files, entries[str] ∈ dirs0, ino ∈ dirs

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)%⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)