Nuprl - Proof: select listify id 1 1 2

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At: select listify id 1 1 2

1. T : Type
2. n :

3. f :

n

T
4. j : {...n-1}
5. 0

j+1

(

i:{(j+1)..n

}. (f{(j+1)..n

})[(i-(j+1))] = f(i))
6. 0

j
7. i : {j..n

}

(f{j..n

})[(i-j)] = f(i)

By:	RecCaseSplit `listify`

Generated subgoals:

1 8. nj
nil[(i-j)] = f(i) T
1 step

2 8. j<n
(f(j).f{(j+1)..n})[(i-j)] = f(i)
5 steps

About:

IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html

(12steps total) PrintForm Definitions Lemmas list 1 Sections StandardLIB Doc