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Proof step 43 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * abs(sdt, entries) * 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), ino1 ∈ dirs, str ∈ entries, str0 ∈ entries, p3 ∈ tree, ino0 ∈ dirs, p1 ∈ sdt[str0], str0 ∈ sdt, p0 ∈ sdt[str], str ∈ sdt, ∀ 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

Proof step 44:
⊦ str ∈ sdt → abs(sdt, sdi) = (⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str]))
Proof step 45:
tree[p3] = dnode(md, sdt), (( m => mkdir(# entries, md, m1, m2, entries) * (⌜str ∈ entries⌝ ∧ abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p3, m)) )(dirs), tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, ispath(p1, entries[str0], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), p0 ∈ sdt[str], str0 ∈ sdt, p1 ∈ sdt[str0], ino0 ∈ dirs, p3 ∈ tree, str0 ∈ entries, str ∈ entries, ino1 ∈ dirs, str ∈ sdt, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

Rule argument

spec: <toplevel> name: entries-recdel seq: ⊦ str ∈ sdt → abs(sdt, sdi) = ( ⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str])) subst: [str, sdi, sdt] → [str, entries, sdt] flags: <all paths>

Additional information

Used terms: str, entries, sdt