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Proof step 2 from proof sep-split-interactive (previous step)

((P * P0))(H) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H0 ∪ H1 = H ∧ P(H0) ∧ P0(H1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 4:
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P(H0) ∧ P0(H1) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H0 ∪ H1 = H ∧ P(H0) ∧ P0(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, P0, P] flags: <all paths>

Additional information

Used terms: H, P0, P