Nuprl - Proof: Ramsey-recursion 1 2 1 1 2 1 1 2 2 1 2 1

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At: Ramsey-recursion 1 2 1 1 2 1 1 2 2 1 2 1

1. r:
2. k:
3. L: List
4. R: List
5. 2k
6. ||R|| = ||L||
7. i:||L||. 0 < L[i] R[i]- > L[i--]^k
8. r- > R^k-1
9. n:
10. r+1n
11. G: {s:(n List)| ||s|| = k & (x,y:||s||. x < y s[x] < s[y]) }||L||
12. c: ||R||
13. f: R[c](n-1)
14. increasing(f;R[c])
15. s:R[c] List. ||s|| = k-1 (x,y:||s||. x < y s[x] < s[y]) G(map(f;s) @ [(n-1)]) = c
16. 0 < L[c]
17. R[c]- > L[c--]^k
18. (s. G(map(f;s))) {s:(R[c] List)| ||s|| = k & (x,y:||s||. x < y s[x] < s[y]) }||L[c--]||
19. c@0: ||L[c--]||
20. f@0: L[c@0]R[c]
21. increasing(f@0;L[c@0])
22. s:L[c@0] List. ||s|| = k (x,y:||s||. x < y s[x] < s[y]) G(map(f;map(f@0;s))) = c@0
23. c@0 ||L||
24. c@0 = c
25. increasing(f o f@0;L[c@0])
26. s: L[c@0] List
27. ||s|| = k (x,y:||s||. x < y s[x] < s[y]) G(map(f;map(f@0;s))) = c@0