Nuprl - Proof: select append back 1 2

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At: select append back 1 2

1. T : Type
2. T List
3. bs : T List
4. u : T
5. v : T List
6.

i:{||v||..(||v||+||bs||)

}. (v @ bs)[i] = bs[(i-||v||)]
7. i : {(||v||+1)..(||v||+1+||bs||)

}

(u.(v @ bs))[i] = bs[(i-(||v||+1))]

By:	RWH (LemmaC Thm* a:T, as:T List, i:. 0<i i\|\|as\|\| (a.as)[i] = as[(i-1)]) 0

Generated subgoal:

1 (v @ bs)[(i-1)] = bs[(i-(||v||+1))]
2 steps

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