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Proof step 3 from proof owns-none-one (previous step)

r ∈ rs, ∀ r. r ∈ rs → g(r) ≠ a ⊦ # rs = 0, r = null, rs = ∅, owns-one(a, r, rs, g(r; a))

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
r ∈ rs, r ∈ rs → g(r) ≠ a, ∀ r. r ∈ rs → g(r) ≠ a ⊦ # rs = 0, r = null, rs = ∅, owns-one(a, r, rs, g(r; a))

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r