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Proof step 100 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), 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 ∈ files, links(ino0, dirs0) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ [c_inode := getinode (mkdentry(str, ino1).ino, dirs, files[ino, files[ino].size:= inode.size][ino, files[ino, files[ino].size:= inode.size][ino].meta:= inode.meta]); asynced := ?] ( afs-cons(dirs, files[ino, files[ino].size:= inode.size][ino, files[ino, files[ino].size:= inode.size][ino].meta:= inode.meta]) → of-cons(of, dirs, files[ino, files[ino].size:= inode.size][ino, files[ino, files[ino].size:= inode.size][ino].meta:= inode.meta])), ino1 = 0, ino1 ∈ dirs0, ino1 ∈ files, str ∈ entries

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)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)