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Proof step 2 from proof disj-alloc (previous step)

H0 ⊥ H1 ⊦ r0 = null, r0 ∈ H0, r0 ∈ H1, H0 ⊥ H1 ++ r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information