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Proof step 69 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H4 ∪ H3, (H1 ++ r)[r] = H2[r], !(O2(r), O2, r =>_)(H0), !(O2(r), O2, P)(H4), r =>_(H0), P(H4), P0(H3), r ∈ H2, r ∈ H4, H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, ∀ r0. ((r = r0 ⊃ O2(r0);O1(r0)) = O2(r) ↔ O2(r0) = O2(r)) ∧ ((r = r0 ⊃ O2(r0);O1(r0)) = O2(r) → (r0 = r ∨ r0 ∈ H1 ↔ r0 ∈ H2) ∧ (r0 ∈ H1 → (H1 ++ r)[r0] = H2[r0])) ⊦ O1(r) = O2(r), ((!(O2(r), O1, P) * P0))(H1), ((!(O2(r), O1, P) * P0))(H4 ∪ H3), r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 70:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)
Proof step 71:
H1 ++ r = H0 ∪ H4 ∪ H3, (H1 ++ r)[r] = H2[r], !(O2(r), O2, P)(H4), !(O2(r), O2, r =>_)(H0), r =>_(H0), P(H4), P0(H3), r ∈ H4, r ∈ H2, H0 ⊥ H3, H4 ⊥ H3, ∀ r0. ((r = r0 ⊃ O2(r0);O1(r0)) = O2(r) ↔ O2(r0) = O2(r)) ∧ ((r = r0 ⊃ O2(r0);O1(r0)) = O2(r) → (r0 = r ∨ r0 ∈ H1 ↔ r0 ∈ H2) ∧ (r0 ∈ H1 → (H1 ++ r)[r0] = H2[r0])), ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H4 ⊦ O1(r) = O2(r), ((!(O2(r), O1, P) * P0))(H4 ∪ H3), ((!(O2(r), O1, P) * P0))(H1), r ∈ H1

Rule argument

spec: heap name: disj seq: ⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) subst: [H1, H0] → [H4, H0] flags: <all paths>

Additional information

Used terms: H4, H0