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Proof step 48 from proof alloc-p (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 49:
¬ r ∈ H, r ≠ null ⊦ ((r =>_ * P))(H ++ r) ↔ P(H)
Proof step 50:
H1 ++ r = H0 ∪ H, ((r =>_ * !(me, O1, P) * P0))(H1 ++ r) ↔ ((!(me, O1, P) * P0))(H1), !(me, O2, r =>_)(H0), ((!(me, O2, P) * P0))(H), rely(me, O1(r; me), H1 ++ r, O2, H2), H0 ⊥ H ⊦ r = null, ((!(me, O1, P) * P0))(H1), r ∈ H1

Rule argument

spec: maplet name: mapex-new-rw seq: ¬ r ∈ H, r ≠ null ⊦ ((r =>_ * P))(H ++ r) ↔ P(H) subst: [r, H, P] → [r, H1, !(me, O1, P) * P0]

Additional information

Used terms: r, H1, !(me, O1, P) * P0