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Proof step 39 from proof alloc-p (previous step)

H1 ++ r = H0 ∪ H4 ∪ H3, P(H4), P0(H3), r0 ∈ H4, H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ H3, ∀ r1. (O1(r; O1(r0))(r1) = O1(r0) ↔ O2(r1) = O1(r0)) ∧ (O1(r; O1(r0))(r1) = O1(r0) → (r1 ∈ H1 ++ r ↔ r1 ∈ H2) ∧ (r1 ∈ H1 ++ r → (H1 ++ r)[r1] = H2[r1])) ⊦ O2(r0) = O1(r0), r = null, r ∈ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
H1 ++ r = H0 ∪ H4 ∪ H3, P(H4), P0(H3), r0 ∈ H4, H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, ∀ r1. ((r = r1 ⊃ O1(r0);O1(r1)) = O1(r0) ↔ O2(r1) = O1(r0)) ∧ ((r = r1 ⊃ O1(r0);O1(r1)) = O1(r0) → (r1 = r ∨ r1 ∈ H1 ↔ r1 ∈ H2) ∧ (r1 = r ∨ r1 ∈ H1 → (H1 ++ r)[r1] = H2[r1])) ⊦ O2(r0) = O1(r0), r = null, r ∈ H1

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)