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At: sum switch 1

1. n : 
2. f : n
3. i : (n-1)
  sum(f((ii+1)(x)) | x < n) = sum(f(x) | x < n)


By: Inst Thm* k:i,j:k. (ij kk [n;i;i+1]
THEN
Subst'
(sum(f((ii+1)(x)) | x < n)
(=
(sum(f((ii+1)(x)) | x < i)+sum(f((ii+1)(x+i)) | x < n-i))
0


Generated subgoals:

1 4. (ii+1)  nn
  sum(f((ii+1)(x)) | x < n)
  =
  sum(f((ii+1)(x)) | x < i)+sum(f((ii+1)(x+i)) | x < n-i)

1 step
2 4. (ii+1)  nn
  sum(f((ii+1)(x)) | x < i)+sum(f((ii+1)(x+i)) | x < n-i)
  =
  sum(f(x) | x < n)

13 steps

About:
intnatural_numberaddsubtractapplyfunctionequalmemberall
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(16steps total) PrintForm Definitions Lemmas mb nat Sections MarkB generic Doc