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Proof step 3 from proof alloc-p (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
((!(me, O1, P) * P0))(H1), rely(me, O1(r; me), H1 ++ r, O2, H2) ⊦ r = null, ((!(me, O2, r =>_ * P) * P0))(H1 ++ r), r ∈ H1
Proof step 41:
((!(me, O2, r =>_ * P) * P0))(H1 ++ r), rely(me, O1(r; me), H1 ++ r, O2, H2) ⊦ r = null, ((!(me, O1, P) * P0))(H1), r ∈ H1

Rule argument

pos: right 1

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