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Proof step 45 from proof alloc-p (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 46:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 47:
rely(me, O1(r; me), H1 ++ r, O2, H2), ∃ H0, H2. H0 ⊥ H2 ∧ H1 ++ r = H0 ∪ H2 ∧ !(me, O2, r =>_)(H0) ∧ ((!(me, O2, P) * P0))(H2) ⊦ r = null, ((!(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] → [H1 ++ r, !(me, O2, P) * P0, !(me, O2, r =>_)] flags: <all paths>

Additional information

Used terms: H1 ++ r, !(me, O2, P) * P0, !(me, O2, r =>_)