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Proof step 3 from proof split-union (previous step)

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

User applied rule insert rewrite 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(H0), P1(H1), H0 ⊥ H1 ⊦ ∃ H2, H3. H2 ⊥ H3 ∧ H0 ∪ H1 = H2 ∪ H3 ∧ P0(H2) ∧ P1(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] → [H0 ∪ H1, P1, P0] flags: <all paths>

Additional information

Used terms: H0 ∪ H1, P1, P0