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Proof step 9 from proof precise-sep (previous step)

H2 ∪ H0 = (H4 ∪ H3) ∪ H, ((P0 * P1))(H2), P0(H4), P1(H3), precise(P1), precise(P0), H4 ⊥ H3, H2 ⊥ H0, H ⊥ H4, H ⊥ H3 ⊦ H2 = H4 ∪ H3

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 11:
H2 ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P1(H3), precise(P0), precise(P1), H ⊥ H3, H ⊥ H4, H2 ⊥ H0, H4 ⊥ H3, ∃ H0, H1. H0 ⊥ H1 ∧ H2 = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1) ⊦ H2 = H4 ∪ H3

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] → [H2, P1, P0] flags: <all paths>

Additional information

Used terms: H2, P1, P0