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Proof step 105 from proof inv-mkdir (previous step)

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

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 106:
⊦ of-cons(of0, dirs0, files0) → (of-cons(of1, dirs1, files1) ↔ (∀ n. (n ∈ of0 → of0[n].ino ∈ files0) → n ∈ of1 → of1[n].ino ∈ files1))

Rule argument

Additional information

Used simplifier rules: of-cons-reduce: ⊦ of-cons(of0, dirs0, files0) → ( of-cons(of1, dirs1, files1) ↔ (∀ n. (n ∈ of0 → of0[n].ino ∈ files0) → n ∈ of1 → of1[n].ino ∈ files1));% ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).meta = md; (s)% ⊦ mkdentry(str, m).dentry?; (s)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)