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Proof step 14 from proof sharing-emp (previous step)

P(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2), ¬ r ∈ H1 ∧ H0 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 16:
P(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2), ¬ r ∈ H1 ∧ (∀ r. (r ∈ H0 ↔ r ∈ ∅) ∧ (r ∈ H0 → H0[r] = (∅)[r]))

Rule argument

spec: heap name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [∅, H0] flags: <all paths>

Additional information

Used terms: ∅, H0