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Proof step 3 from proof max-plus-one (previous step)

ns.max + 1 ∈ ns ⊦ # ns = 0, ns = ∅

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ n ∈ ns → ¬ ns.max < n
Proof step 5:
ns.max + 1 ∈ ns, ns.max + 1 ∈ ns → ¬ ns.max < ns.max + 1 ⊦ # ns = 0, ns = ∅

Rule argument

spec: oset, inst: nat-set name: Max-max seq: ⊦ n ∈ ns → ¬ ns.max < n subst: [n, ns] → [ns.max + 1, ns]

Additional information

Used terms: ns.max + 1, ns