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Proof step 29 from proof alloc-p (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id)
Proof step 31:
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, !(me, O2, P)(H4), r ∈ H1

Rule argument

spec: <toplevel> name: owned seq: ⊦ !(id, O, P)(H) ↔ P(H) ∧ (∀ r. r ∈ H → O(r) = id) subst: [H, P, O, id] → [H4, P, O1, me] flags: <all paths>

Additional information

Used terms: H4, P, O1, me