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Proof step 34 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H4 ∪ H3, P(H4), P0(H3), rely(me, O1(r; me), H1 ++ r, O2, H2), H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, ∀ r. r ∈ H4 → O1(r) = me ⊦ r = null, r ∈ H1, P(H4) ∧ (∀ r. r ∈ H4 → O2(r) = me)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
H1 ++ r = H0 ∪ H4 ∪ H3, P(H4), P0(H3), rely(me, O1(r; me), H1 ++ r, O2, H2), r0 ∈ H4, H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ H3, ∀ r. r ∈ H4 → O1(r) = me ⊦ O2(r0) = me, r = null, r ∈ H1

Proof step 34 from proof alloc-p (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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