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Proof step 3 from proof precise-disj (previous step)

H0 ∪ H2 = H1 ∪ H3, ((P0 * P1 * true))(H1 ∪ H3), P0(H0), P1(H1), precise(P0), precise(P1), H1 ⊥ H3, H0 ⊥ H2 ⊦ H0 ⊥ H1

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 5:
((P0 * P1 * true))(H1 ∪ H3) ↔ (∃ H0, H2. H0 ⊥ H2 ∧ H1 ∪ H3 = H0 ∪ H2 ∧ P0(H0) ∧ ((P1 * true))(H2)), H0 ∪ H2 = H1 ∪ H3, ((P0 * P1 * true))(H1 ∪ H3), P0(H0), P1(H1), precise(P1), precise(P0), H0 ⊥ H2, H1 ⊥ H3 ⊦ H0 ⊥ H1

Rule argument

spec: <toplevel> name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [P0, P1, H] → [P0, P1 * true, H1 ∪ H3]

Additional information

Used terms: P0, P1 * true, H1 ∪ H3