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Proof step 3 from proof owns-none-one2 (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 1 terms: r

Additional information

Used terms: r