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Proof step 12 from proof exin-uni (previous step)

⊦ (H0 ++ r)[r, o] = H0 ∪ (∅ ++ r)[r, o], r = null, r ∈ H0, r ≠ null ∧ r ∈ ∅ ++ r ∧ H0 ⊥ ∅ ++ r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ (H0 ++ r)[r, o] = H0 ∪ (∅ ++ r)[r, o], r = null, r ∈ H0, H0 ⊥ ∅ ++ r

Rule argument

Additional information

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