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Proof step 49 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), (( m => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str -- str0, entries -- str -- str0) * abs(sdt[str0], entries[str0])) * abs(sdt[str], entries[str]) * abs|(tree, ino0, p3, m)) )(dirs), tree[p3 + str0 ' + p1].dir?, tree[p3 + str ' + p0].dir?, ispath(p0, entries[str], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p1, entries[str0], ino1, dirs, files), str ∈ sdt, ino1 ∈ dirs, str ∈ entries, str0 ∈ entries, p3 ∈ tree, ino0 ∈ dirs, p1 ∈ sdt[str0], str0 ∈ sdt, p0 ∈ sdt[str], ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ str = str0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: path-unfold seq: ⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)) subst: [p, ino0, tree] → [p0, entries[str], sdt[str]] flags: <all paths>

Additional information

Used terms: p0, entries[str], sdt[str]