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Proof step 111 from proof inv-create (previous step)

dirs0[ino0] = mkdir(m2, md1, # links(ino0, dirs0), m4, entries), dirs = dirs0[ino0, mkdir(m2 + 1, md1, # links(ino0, dirs0), m4, entries[str, ino1])], files = (files0 ++ ino1)[ino1, mkfile(0, md, 1, ∅)], ino = inode.ino, inode = mkinode(ino0, md1, true, # links(ino0, dirs0), m4, m2), afs-cons(dirs0, files0), ispath(q, ROOT_INO, ino0, dirs0, files0), of-cons(of, dirs0, files0), pw(user, md1), px(user, md1), px-parent(user, q, ROOT_INO, dirs0, files0), ino0 ∈ dirs0, links(ino0, dirs0) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ [c_inode := getinode(mkdentry(str, ino1).ino, dirs, files); asynced := ?] (afs-cons(dirs, files) → of-cons(of, dirs, files)), ino1 = 0, ino ∈ files, str ∈ entries, ino1 ∈ files0, ino1 ∈ dirs0, ino ∈ dirs

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkdentry(str, m).dentry?; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)%ino ≠ 0, ¬ ino ∈ files ⊦ ino0 ∈ files ++ ino ↔ ino0 = ino ∨ ino0 ∈ files; (s)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)