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Proof step 1 from proof ex-p (conclusion)

⊦ ((_∃ xP * P))(H) ↔ (∃ x. ((xP(x) * P))(H))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 3:
⊦ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (_∃ xP)(H0) ∧ P(H1)) ↔ (∃ x. ((xP(x) * P))(H))

Rule argument

spec: heap-sep name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [H, P, _∃ xP] flags: <all paths>

Additional information

Used terms: H, P, _∃ xP