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Proof step 47 from proof abs-acyclic (previous step)

tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, sdt[str] = dnode(md, sdt0), (( ( ( (ROOT_INO => mkdir(# entries0, md, m0, m1, entries0) * abs(sdt0, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), sdt[str].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ sdt, str ∈ entries ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, sdt[str] = dnode(md, sdt0), (( ( ( (ROOT_INO => mkdir(# entries0, md, m0, m1, entries0) * abs(sdt0, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ sdt, str ∈ entries ⊦

Rule argument

Additional information

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