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Proof step 32 from proof abs-nlinks-h (previous step)

tree[p0] = dnode(md, sdt), tree[p0 + str '] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, ((m0 => mkdir(# entries, md, m1, m2, entries) * abs(sdt, entries) * abs|(tree, ino0, p0, m0)))(dirs), ispath(p0, ino0, m0, dirs, files), p0 ∈ tree, ino1 ∈ files, ino0 ∈ dirs, str ∈ entries, str ∈ sdt, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, entries[str] ∈ dirs ∨ entries[str] ∈ files ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 33:
⊦ str ∈ sdt → abs(sdt, sdi) = (⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str]))
Proof step 34:
tree[p0 + str '] = fnode(⌜ino1⌝ ), tree[p0] = dnode(md, sdt), links(ino1, dirs) = ∅, (( m0 => mkdir(# entries, md, m1, m2, entries) * (⌜str ∈ entries⌝ ∧ abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p0, m0)) )(dirs), ispath(p0, ino0, m0, dirs, files), str ∈ entries, ino0 ∈ dirs, ino1 ∈ files, p0 ∈ tree, str ∈ sdt, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, entries[str] ∈ dirs ∨ entries[str] ∈ files ⊦

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