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Proof step 6 from proof alloc (previous step)

P(H), ∃ r. ¬ r ∈ H ∧ r ≠ null ⊦ ∃ r0. ¬ r0 ∈ H ∧ r0 ≠ null

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information