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Proof step 52 from proof alloc-p (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 53:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 54:
H1 ++ r = H0 ∪ H, !(me, O2, r =>_)(H0), ((!(me, O2, P) * P0))(H), rely(me, O1(r; me), H0 ∪ H, O2, H2), H0 ⊥ H ⊦ r = null, ((!(me, O1, P) * P0))(H1), ((r =>_ * !(me, O1, P) * P0))(H0 ∪ H), r ∈ H1, r =>_(H0) ∧ ((!(me, O1, P) * P0))(H) ∧ H0 ⊥ H

Rule argument

spec: heap-sep name: split-union seq: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1) subst: [H1, H0, P1, P0] → [H, H0, !(me, O1, P) * P0, r =>_] flags: <all paths>

Additional information

Used terms: H, H0, !(me, O1, P) * P0, r =>_