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Proof step 18 from proof sep-ass (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ H0 ⊥ H1 → H1 ∪ H0 = H0 ∪ H1

Rule argument

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

Additional information

Used simplifier rules: ⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H); (ws)% ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H0, H