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Proof step 28 from proof sharing-com (previous step)

P0(H1 ∪ H0), P1(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ H2 ∪ H1 ∪ H0 = (H1 ∪ H2) ∪ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ H0 ⊥ H1 → H1 ∪ H0 = H0 ∪ H1
Proof step 30:
P0(H1 ∪ H0), P1(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H0, H2 ⊥ H1 ⊦ H2 ∪ H1 ∪ H0 = (H2 ∪ H1) ∪ H0

Rule argument

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

Additional information

Used terms: H2, H1