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Proof step 6 from proof below-nonempty (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
⊦ # ns = 0, n = 0, ns = ∅, ns.min ∈ below(ns, ns.min + n)

Rule argument

pos: left 1 terms: ns.min abort: F discard: T computed: F

Additional information

Used terms: ns.min