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Proof step 20 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H4 ∪ H3, r =>_ = !(me, O2, r =>_), !(me, O1, P)(H4), r =>_(H0), P0(H3), rely(me, O1(r; me), H0 ∪ H4 ∪ H3, O2, H2), H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3 ⊦ r = null, (((!(me, O2, r =>_) * !(me, O2, P)) * P0))(H0 ∪ H4 ∪ H3), r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ (P0 * P1) * P2 = P0 * P1 * P2
Proof step 22:
H1 ++ r = H0 ∪ H4 ∪ H3, r =>_ = !(me, O2, r =>_), !(me, O1, P)(H4), r =>_(H0), P0(H3), rely(me, O1(r; me), H0 ∪ H4 ∪ H3, O2, H2), H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ H3 ⊦ r = null, ((!(me, O2, r =>_) * !(me, O2, P) * P0))(H0 ∪ H4 ∪ H3), r ∈ H1

Rule argument

spec: heap-sep name: sep-ass seq: ⊦ (P0 * P1) * P2 = P0 * P1 * P2 subst: [P2, P1, P0] → [P0, !(me, O2, P), !(me, O2, r =>_)] flags: <all paths>

Additional information

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