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

((oP(o) * P0))(H1), P(H0), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ∃ o. ((!(me, O, oP(o) * P0) * P))(H1 ∪ H0)

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:
P(H0), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ oP(o)(H0) ∧ P0(H2) ⊦ ∃ o. ((!(me, O, oP(o) * P0) * P))(H1 ∪ H0)

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] → [H1, P0, oP(o)] flags: <all paths>

Additional information

Used terms: H1, P0, oP(o)