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Proof step 30 from proof sharing-com (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ 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] → [H0, H1, H2] flags: <all paths>

Additional information

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