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Proof step 9 from proof sep-ass (previous step)

P0(H2), P1(H), P2(H0), H2 ⊥ H, H0 ⊥ H2, H0 ⊥ H ⊦ ((P0 * P1 * P2))((H2 ∪ H) ∪ H0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H
Proof step 11:
P0(H2), P1(H), P2(H0), H0 ⊥ H, H2 ⊥ H, H ⊥ H0, H0 ⊥ H2 ⊦ ((P0 * P1 * P2))(H2 ∪ H ∪ H0)

Rule argument

spec: heap name: uni-ass seq: ⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H subst: [H, H1, H0] → [H0, H, H2] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H0, H, H2