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Proof step 16 from proof abs-unfold (previous step)

tree[p] = dnode(md, sdt0), ((ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt0, entries) * abs|(tree, ROOT_INO, p, ino)))(dirs), tree[p].dir?, afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), p ∈ tree ⊦ abs(tree0, ROOT_INO)(dirs0), ∃ sdt. ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(dirs) ∧ tree[p] = dnode(md, sdt)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
tree[p] = dnode(md, sdt0), ((ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt0, entries) * abs|(tree, ROOT_INO, p, ino)))(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), p ∈ tree ⊦ abs(tree0, ROOT_INO)(dirs0), ∃ sdt. ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(dirs) ∧ dnode(md, sdt0) = dnode(md, sdt)

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt).dir?; (s)