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

VFS_PAGE_SIZE = n0 + m % VFS_PAGE_SIZE, files1[ino2] = mkfile(m, md, # links(ino2, dirs), content0), files = files0, files0 = files1 [ino2, mkfile(m, md, # links(ino2, dirs), content0[m / VFS_PAGE_SIZE, fill(content0[m / VFS_PAGE_SIZE], zero, m % VFS_PAGE_SIZE, n0)] upto m) ], ino = inode.ino, inode = mkinode(ino2, md, false, # links(ino2, dirs), 0, n), inode0 = mkinode(ino2, md, false, # links(ino2, dirs), 0, m), afs-cons(dirs, files1), ispath(p, ROOT_INO, ino2, dirs, files1), of-cons(of, dirs, files1), pw(user, md), px-parent(user, p, ROOT_INO, dirs, files1), m / VFS_PAGE_SIZE ∈ content0, ino2 ∈ files1, m ≤ n ⊦ [asynced := ?] (afs-cons(dirs, files) → of-cons(of, dirs, files)), m % VFS_PAGE_SIZE = 0, ino ∈ dirs, ino2 ∈ dirs, ino ∈ files

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)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)%⊦ afs-cons(dirs, files) ∧ ¬ ino ∈ dirs ∧ ¬ ino ∈ files → links(ino, dirs) = ∅; (s)% ⊦ # ∅ = 0; (s)