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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:kx,y:||p(j)||. x<y  (p(j))[x]>(p(j))[y]
8. j:kx:||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).
Thm* (i:nf(i) = g(i))  sum(f(x) | x < n) = sum(g(x) | x < n)
THEN
SplitOnConclITE


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About:
listconsifthenelseintnatural_number
subtractless_thanapplyfunctionequalimpliesall
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