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Proof step 2 from proof owns-none-one2 (previous step)

owns-none(a, rs, g) ⊦ r = null, owns-one(a, r, rs ++ r, g(r; a)), r ∈ rs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ r. r ∈ rs → g(r) ≠ a ⊦ r = null, owns-one(a, r, rs ++ r, g(r; a)), r ∈ rs

Rule argument

Additional information

Used simplifier rules: owns-none: ⊦ owns-none(a, rs, g) ↔ (∀ r. r ∈ rs → g(r) ≠ a);