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Proof step 8 from proof del-new (previous step)

r0 ∈ H, r0 ∈ H ++ r ⊦ (H ++ r)[r0] = H[r0], r = null, r0 = r, r ∈ H

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
r0 ∈ H ⊦ (H ++ r)[r0] = H[r0], r = null, r ∈ H

Rule argument

Additional information

Used simplifier rules: in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;