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Proof step 13 from proof sharing-com (previous step)

P0(H2 ∪ H0), P1(H1 ∪ H0), H2 ⊥ H0, H1 ⊥ H0, H2 ⊥ H1 ⊦ H2 ∪ H1 ∪ H0 = (H1 ∪ H2) ∪ H0, ∃ H5, H4, H3. H5 ⊥ H3 ∧ H5 ⊥ H4 ∧ H4 ⊥ H3 ∧ H2 ∪ H1 ∪ H0 = H5 ∪ H4 ∪ H3 ∧ P1(H5 ∪ H3) ∧ P0(H4 ∪ H3)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ H0 ⊥ H1 → H1 ∪ H0 = H0 ∪ H1
Proof step 15:
P0(H2 ∪ H0), P1(H1 ∪ H0), H1 ⊥ H0, H2 ⊥ H0, H2 ⊥ H1 ⊦ H2 ∪ H1 ∪ H0 = (H2 ∪ H1) ∪ H0, ∃ H5, H4, H3. H5 ⊥ H3 ∧ H5 ⊥ H4 ∧ H4 ⊥ H3 ∧ H2 ∪ H1 ∪ H0 = H5 ∪ H4 ∪ H3 ∧ P1(H5 ∪ H3) ∧ P0(H4 ∪ H3)

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