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Proof step 22 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), H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ 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 23:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 24:
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, !(me, O2, r =>_)(H0) ∧ ((!(me, O2, P) * P0))(H4 ∪ H3) ∧ H0 ⊥ H4 ∪ H3

Rule argument

spec: heap-sep name: split-union seq: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1) subst: [H1, H0, P1, P0] → [H4 ∪ H3, H0, !(me, O2, P) * P0, !(me, O2, r =>_)] flags: <all paths>

Additional information

Used terms: H4 ∪ H3, H0, !(me, O2, P) * P0, !(me, O2, r =>_)