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Proof step 118 from proof inv-link (previous step)

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 = EACCESS, err1 = err2, err2 = ESUCCESS, ino = ino1, inode = getinode(ino, dirs, files), inode0 = mkinode(entries0[str], md, false, # links(entries0[str], dirs), 0, m), inode.directory, 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-iget(ino, dirs, files), px(user, md0), px-parent(user, q, ROOT_INO, dirs, files), px-parent(user, parent(p), ROOT_INO, dirs, files), str ∈ entries0, entries0[str] ∈ files, ino0 ∈ dirs, ¬ px(user, inode.meta) ∨ ¬ pw(user, inode.meta), 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: ⊦ pre-iget(ino, dirs, files) ↔ valid-ino(ino, dirs, files); (s)% ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (s)%⊦ getinode(ino, dirs, files) = (ino ∈ dirs ⊃ mkinode (ino, dirs[ino].meta, true, dirs[ino].nlink, dirs[ino].nsubdirs, dirs[ino].size) ;mkinode (ino, files[ino].meta, false, files[ino].nlink, 0, files[ino].size) ); (s)