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Proof step 21 from proof abs-unfold (previous step)

tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), p ∈ tree ⊦ (_∃(λ ino1. abs|(tree, ROOT_INO, p, ino1) * abs(tree[p], ino1)))(dirs), abs(tree0, ROOT_INO)(dirs0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(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), ∃ ino. ((abs|(tree, ROOT_INO, p, ino) * abs(dnode(md, sdt), ino)))(dirs)

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ nD)(dirs) ↔ (∃ ino. nD(ino)(dirs)); (s)