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Proof step 10 from proof entries-unfold (previous step)

((abs(sdt, entries[str, ino]) * D))(dirs), str ∈ sdt ⊦ str ∈ entries, ∃ tree, ino0, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ entries[str, ino] = sdi0[str, ino0] ∧ ((abs(tree, ino0) * abs(sdt0, sdi0) * D))(dirs)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ str ∈ sdt → abs(sdt, sdi) = (⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str]))
Proof step 12:
(((⌜str ∈ entries[str, ino]⌝ ∧ abs(sdt -- str, entries[str, ino] -- str) * abs(sdt[str], entries[str, ino][str])) * D))(dirs), str ∈ sdt ⊦ str ∈ entries, ∃ tree, ino0, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ entries[str, ino] = sdi0[str, ino0] ∧ ((abs(tree, ino0) * abs(sdt0, sdi0) * D))(dirs)

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[str, ino], sdt] flags: <all paths>

Additional information

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