Nuprl - Proof: iter via intseg notnull 1

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At: iter via intseg notnull 1

1. A : Type
2. f : A

A
3. u : A
4. a :

5. b :

6. e : {a..b

}

A
7. a<b
8. c :

9. b = c+1

(Iter(f;u) i:{a..b

}. e(i)) = f((Iter(f;u) i:{a..c

}. e(i)),e(c))

By:	Compute Iter(f;u) i:{a..b}. e(i) THEN SplitITE 0

Generated subgoal:

1 10. a<b
f((Iter(f;u) i:{a..b-1}. e(i)),e(b-1)) = f((Iter(f;u) i:{a..c}. e(i)),e(c))
1 step

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IF YOU CAN SEE THIS go to http://www.cs.cornell.edu/Info/People/sfa/Nuprl/Shared/Xindentation_hack_doc.html