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

((P -* P0))(H1), ((P -* P1))(H2), P(H), H2 ⊥ H, H1 ⊥ H2, H1 ⊥ H ⊦ ((P0 ⊗ P1))(H1 ∪ H2 ∪ H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2))
Proof step 32:
((P -* P1))(H2), ((P -* P0))(H1), P(H), H1 ⊥ H, H1 ⊥ H2, H2 ⊥ H ⊦ ∃ H0, H3, H4. H0 ⊥ H4 ∧ H0 ⊥ H3 ∧ H3 ⊥ H4 ∧ H1 ∪ H2 ∪ H = H0 ∪ H3 ∪ H4 ∧ P0(H0 ∪ H4) ∧ P1(H3 ∪ H4)

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] → [H1 ∪ H2 ∪ H, P1, P0] flags: <all paths>

Additional information

Used terms: H1 ∪ H2 ∪ H, P1, P0