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Proof step 55 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H, !(me, O2, r =>_)(H0), ((!(me, O2, P) * P0))(H), r =>_(H0), rely(me, O1(r; me), H1 ++ r, O2, H2), H0 ⊥ H ⊦ r = null, ((!(me, O1, P) * P0))(H1), ((!(me, O1, P) * P0))(H), r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 56:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 57:
H1 ++ r = H0 ∪ H, !(me, O2, r =>_)(H0), r =>_(H0), rely(me, O1(r; me), H1 ++ r, O2, H2), H0 ⊥ H, ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ !(me, O2, P)(H0) ∧ P0(H1) ⊦ r = null, ((!(me, O1, P) * P0))(H), ((!(me, O1, P) * P0))(H1), r ∈ 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, P0, !(me, O2, P)] flags: <all paths>

Additional information

Used terms: H, P0, !(me, O2, P)