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Proof step 41 from proof inv-mkdir (previous step)

dirs0[m] = mkdir(m0, md2, # links(m, dirs0), m2, entries0), dirs = (dirs0[m, mkdir(m0 + 1, md2, # links(m, dirs0), m2 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)], ino = inode.ino, inode = mkinode(m, md0, true, # links(m, dirs0), m2, m0), afs-cons(dirs0, files), afs-cons((dirs0[m, mkdir(m0 + 1, md2, # links(m, dirs0), m2 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)], files), m ∈ dirs0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ [skip] ( afs-cons(dirs, files) → ⟪p_inode := (m ∈ dirs ⊃ mkinode(m, dirs[m].meta, true, dirs[m].nlink, dirs[m].nsubdirs, dirs[m].size) ;mkinode(m, files[m].meta, false, files[m].nlink, 0, files[m].size) ); c_inode := getinode(mkdentry(str, ino0).ino, dirs, files); asynced := ?⟫ true), ino0 = 0, ino ∈ files, str ∈ entries0, ino0 ∈ files, ino0 ∈ dirs0, ino ∈ dirs

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

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