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Proof step 11 from proof abs-acyclic (previous step)

abs(tree, ROOT_INO)(dirs) ↔ (_∃(λ ino1. abs|(tree, ROOT_INO, p, ino1) * abs(tree[p], ino1)))(dirs), dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m0, entries), entries[str] = ROOT_INO, tree[p].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 ∈ entries, ino ∈ dirs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
entries[str] = ROOT_INO, dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m0, entries), ((abs|(tree, ROOT_INO, p, ino0) * abs(tree[p], ino0)))(dirs), tree[p].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs, str ∈ entries ⊦

Rule argument

Additional information

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