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Proof step 114 from proof inv-truncate (previous step)

files0[ino2] = mkfile(m, md, # links(ino2, dirs), content0), files = files0[ino2, mkfile(m, md, # links(ino2, dirs), content0 upto min(n, 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, files0), ispath(p, ROOT_INO, ino2, dirs, files0), of-cons(of, dirs, files0), pw(user, md), px-parent(user, p, ROOT_INO, dirs, files0), ino2 ∈ files0 ⊦ [if ino ∈ files then {files[ino].size := inode.size; files[ino].meta := inode.meta} else if ino ∈ dirs then dirs[ino].meta := inode.meta; inode0 := (inode0.ino ∈ dirs ⊃ mkinode(inode0.ino, dirs[inode0.ino].meta, true, dirs[inode0.ino].nlink, dirs[inode0.ino].nsubdirs, dirs[inode0.ino].size) ;mkinode(inode0.ino, files[inode0.ino].meta, false, files[inode0.ino].nlink, 0, files[inode0.ino].size) ); err := ESUCCESS; asynced := ?] (afs-cons(dirs, files) → of-cons(of, dirs, files)), ino2 ∈ dirs, m ≤ n ∧ m % VFS_PAGE_SIZE ≠ 0 ∧ m / VFS_PAGE_SIZE ∈ content0

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

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