Nuprl - Proof: cons sublist cons 2 1 2 2 2 1

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At: cons sublist cons 2 1 2 2 2 1

1. T : Type
2. x1 : T
3. x2 : T
4. L1 : T List
5. L2 : T List
6. x1 = x2
7. f :

||L1||

||L2||
8. increasing(f;||L1||)
9.

||L1||. L1[j] = L2[(f(j))]
10. j :

(||L1||+1)
11.

j = 0

L1[(j-1)] = L2[(f(j-1)+1-1)]

By:	Subst ((f(j-1)+1-1) ~ (f(j-1))) 0

Generated subgoal:

1 L1[(j-1)] = L2[(f(j-1))]
1 step

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