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Proof step 6 from proof sep-ass (previous step)

((P0 * P1))(H1), P2(H0), H1 ⊥ H0 ⊦ ((P0 * P1 * P2))(H1 ∪ H0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 8:
P2(H0), H1 ⊥ H0, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ P0(H0) ∧ P1(H2) ⊦ ((P0 * P1 * P2))(H1 ∪ H0)

Rule argument

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

Additional information

Used terms: H1, P1, P0