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Proof step 10 from proof precise-disj (previous step)

H1 ∪ H3 = H ∪ H4, ((P1 * true * P0))(H ∪ H4), ((P1 * true))(H4), P0(H), P1(H1), precise(P0), precise(P1), H1 ⊥ H3, H ⊥ H4 ⊦ H ⊥ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 12:
H1 ∪ H3 = H ∪ H4, ((P1 * true))(H4), P0(H), P1(H1), precise(P1), precise(P0), H ⊥ H4, H1 ⊥ H3, ∃ H0, H1. H0 ⊥ H1 ∧ H ∪ H4 = H0 ∪ H1 ∧ P1(H0) ∧ ((true * P0))(H1) ⊦ H ⊥ H1

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] → [H ∪ H4, true * P0, P1] flags: <all paths>

Additional information

Used terms: H ∪ H4, true * P0, P1