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Proof step 26 from proof sep-ass (previous step)

P0(H1), P1(H2), P2(H), H2 ⊥ H, H1 ⊥ H2, H1 ⊥ H ⊦ (((P0 * P1) * P2))(H1 ∪ H2 ∪ H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H
Proof step 28:
P0(H1), P1(H2), P2(H), H1 ⊥ H, H1 ⊥ H2, H2 ⊥ H, H ⊥ H1 ⊦ (((P0 * P1) * P2))((H1 ∪ H2) ∪ H)

Rule argument

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

Additional information

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