∵ PMI ↔ LNNP, we proove by suppose LNNP holds.
Let S={n∈N:1∈S:{1,2,3…n}⊆S⟹n+1∈S} Suppose N−S≠∅ By LNNP ∃m∈N−S,m≠1 is the smallest member. Therefore m−1∈S Furthermore m−2,m−3…1∈S ∵1,2,3…m−1∈S⟹m∈S Thus we have a contradiction. Hence S=N
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