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Proof step 8 from proof sharing-emp (previous step)

H1 ∪ H0 = ∅, P(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ P(H2), H1 = ∅ ∧ H0 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 10:
P(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0, ∀ r. (r ∈ H1 ∪ H0 ↔ r ∈ ∅) ∧ (r ∈ H1 ∪ H0 → (H1 ∪ H0)[r] = (∅)[r]) ⊦ P(H2), H1 = ∅ ∧ H0 = ∅

Rule argument

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

Additional information

Used terms: ∅, H1 ∪ H0