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Proof step 7 from proof owns-none-one2 (previous step)

∀ r. r ∈ rs → g(r) ≠ a ⊦ r = null, r ∈ rs, r ≠ null ∧ r ∈ rs ++ r ∧ g(r; a)(r) = a ∧ (∀ r0. r0 ∈ rs ++ r ∧ r0 ≠ r → g(r; a)(r0) ≠ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
r0 ∈ rs, ∀ r. r ∈ rs → g(r) ≠ g(r0) ⊦ # rs = 0, r = null, rs = ∅, r ∈ rs

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]% ⊦ r ≠ r0 → (r ∈ rs ++ r0 ↔ r ∈ rs); (s)% ⊦ ¬ r ∈ rs → (r ∈ rs ++ r0 ↔ r = r0); (s)