Nuprl - Proof: cons sublist cons 1 1 2 1 3 1 1

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At: cons sublist cons 1 1 2 1 3 1 1

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

(||L1||+1)

(||L2||+1)
7. increasing(f;||L1||+1)
8.

(||L1||+1). [x1 / L1][j] = [x2 / L2][(f(j))]
9. [x1 / L1][0] = [x2 / L2][(f(0))]
10. f(0)

0
11. j :

(||L1||+1)
12. [x1 / L1][j] = [x2 / L2][(f(j))]
13. f(0)+j

f(j)

[x1 / L1][j] = L2[(f(j)-1)]

By:	HypSubst -2 0

Generated subgoal:

1 [x2 / L2][(f(j))] = L2[(f(j)-1)]
1 step

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