Nuprl - Proof: finite-partition 2 1 1

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At: finite-partition 2 1 1

1. n:
2. 0 < n
3. k:
4. c: nk
5. p: k( List)
6. sum(||p(j)|| | j < k) = n-1
7. j:k, x,y:||p(j)||. x < y (p(j))[x] > (p(j))[y]
8. j:k, x:||p(j)||. (p(j))[x] < n-1 & c((p(j))[x]) = j

sum(||if c(n-1)=j [(n-1) / (p(j))] else p(j) fi|| | j < k) = sum(if c(n-1)=j ||[(n-1) / (p(j))]|| else ||p(j)|| fi | j < k)

By: BackThru Thm* n:, f,g:(n). (i:n. f(i) = g(i)) sum(f(x) | x < n) = sum(g(x) | x < n) THEN SplitOnConclITE

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(26steps total) PrintForm Definitions Lemmas graph 1 2 Sections Graphs Doc