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Proof step 75 from proof inv-truncate (previous step)

files[ino2] = mkfile(m, md, # links(ino2, dirs), content0), buf-opt = none, content = files[inode.ino].content, err = ESUCCESS, ino = inode.ino, ino0 = min(n, inode.size), ino1 = inode.size % VFS_PAGE_SIZE, ino3 = 0, ino4 = 0, ino5 = ino3, ino6 = ino4, inode = mkinode(ino2, md, false, # links(ino2, dirs), 0, m), pageno = m / VFS_PAGE_SIZE, afs-cons(dirs, files), ispath(p, ROOT_INO, ino2, dirs, files), of-cons(of, dirs, files), pw(user, md), px-parent(user, p, ROOT_INO, dirs, files), ino2 ∈ files ⊦ [if inode.size ≤ n ∧ ino1 ≠ 0 ∧ (pageno ∈ content ∨ buf-opt.some?) then {if buf-opt.some? then buf := buf-opt.buf else buf := files[ino].content[pageno]; buf := fill(buf, zero, ino1, VFS_PAGE_SIZE - ino1); content[pageno] := buf; buf-opt := some(buf)}; files[ino].content := content upto ino0; let new_inode = inode in {new_inode.size := n; afs_inode_sync#(new_inode; dirs, files)}; inode := getinode(inode.ino, dirs, files); err := ESUCCESS; asynced := ?] (afs-cons(dirs, files) → of-cons(of, dirs, files)), VFS_PAGE_SIZE = 0, ino2 ∈ dirs

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

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