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Proof step 7 from proof owns-none-one (previous step)

r ∈ rs, ∀ r. r ∈ rs → g(r) ≠ a ⊦ # rs = 0, g(r) = a, r = null, rs = ∅, r ≠ null ∧ r ∈ rs ∧ g(r; a)(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r; a)(r0) ≠ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information