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Proof step 1 from proof entries-empty (conclusion)

⊦ abs(sdt, ∅) = (⌜sdt = ∅⌝ ∧ emp)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ D = D0 ↔ (∀ D1, dirs. ((D * D1))(dirs) ↔ ((D0 * D1))(dirs))
Proof step 3:
⊦ ∀ D1, dirs. ((abs(sdt, ∅) * D1))(dirs) ↔ (((⌜sdt = ∅⌝ ∧ emp) * D1))(dirs)

Rule argument

spec: heap-sep, inst: dir-heap name: Eq-P seq: ⊦ D = D0 ↔ (∀ D1, dirs. ((D * D1))(dirs) ↔ ((D0 * D1))(dirs)) subst: [D0, D] → [⌜sdt = ∅⌝ ∧ emp, abs(sdt, ∅)] flags: <all paths>

Additional information

Used terms: ⌜sdt = ∅⌝ ∧ emp, abs(sdt, ∅)