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Proof step 5 from proof max-plus-one (previous step)

ns.max + 1 ∈ ns, ns.max + 1 ∈ ns → ¬ ns.max < ns.max + 1 ⊦ # ns = 0, ns = ∅

User applied rule case distinction

And obtained the following new goals

Proof step 6:
ns.max + 1 ∈ ns ⊦ # ns = 0, ns = ∅, ns.max < ns.max + 1

Rule argument

pos: left 2

Additional information