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Proof step 6 from proof sharing-com (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2))
Proof step 8:
P0(H2 ∪ H0), P1(H1 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ 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: <toplevel> name: sharing seq: ⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2)) subst: [H, P1, P0] → [H2 ∪ H1 ∪ H0, P0, P1] flags: <all paths>

Additional information

Used terms: H2 ∪ H1 ∪ H0, P0, P1