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Proof step 8 from proof ex-p (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 10:
∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ xP(x)(H0) ∧ P(H1) ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (∃ x. xP(x)(H0)) ∧ P(H1)

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(x)] flags: <all paths>

Additional information

Used terms: H, P, xP(x)