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Proof step 5 from proof owns-one-none (previous step)

g(r) = g(r0), r0 ∈ rs, r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, g(r0) = a0, r = r0, r = null, rs = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: r0

Additional information

Used terms: r0