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Proof step 15 from proof precise-disj (previous step)

H ∪ H4 = H2 ∪ H0, ((P1 * true))(H4), ((P0 * true))(H0), P0(H), P1(H2), precise(P1), precise(P0), H ⊥ H4, H2 ⊥ H0 ⊦ H ⊥ H2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 17:
H ∪ H4 = H2 ∪ H0, ((P1 * true))(H4), P0(H), P1(H2), precise(P0), precise(P1), H2 ⊥ H0, H ⊥ H4, ∃ H2, H1. H2 ⊥ H1 ∧ H0 = H2 ∪ H1 ∧ P0(H2) ∧ true(H1) ⊦ H ⊥ H2

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] → [H0, true, P0] flags: <all paths>

Additional information

Used terms: H0, true, P0