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Proof step 2 from proof max-plus-one (previous step)

⊦ ns.max + 1 ∈ ns ↔ false, ns = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ns.max + 1 ∈ ns ⊦ # ns = 0, ns = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]