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Proof step 95 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), ino ∈ files, m / VFS_PAGE_SIZE ∈ content0, ino2 ∈ files1, m ≤ n ⊦ [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])), m % VFS_PAGE_SIZE = 0, ino2 ∈ dirs

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

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 ∈ files[ino0, af] ↔ ino ∈ files; (s)% ⊦ files[ino, af][ino0] = (ino0 = ino ∧ ino ∈ files ⊃ af;files[ino0]); (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).size = m2; (s)% ⊦ mkfile(m, md, m0, content).size:= m1 = mkfile(m1, md, m0, content); (s)% ⊦ files[ino, af][ino, af0] = files[ino, af0]; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).meta = md; (s)% ⊦ mkfile(m, md, m0, content).meta:= md0 = mkfile(m, md0, m0, content); (s)