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Proof step 5 from proof below-nonempty (previous step)

∀ n0. n0 ∈ below(ns, ns.min + n) ↔ n0 ∈ ∅ ⊦ # ns = 0, n = 0, ns = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n0. ¬ n0 ∈ below(ns, ns.min + n) ⊦ # ns = 0, n = 0, ns = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ∅; (s)