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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r ∈ rs, ∀ r. r ∈ rs → g(r) ≠ a ⊦ # rs = 0, r = null, rs = ∅, owns-one(a, r, rs, g(r; a))

Rule argument

Additional information

Used simplifier rules: owns-none: ⊦ owns-none(a, rs, g) ↔ (∀ r. r ∈ rs → g(r) ≠ a);%size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]