Nuprl - Proof: pigeon-hole 1 1 1 1

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At: pigeon-hole 1 1 1 1

1. n:
2. m:
3. f: nm
4. Inj(n; m; f)
5. p: m( List)
6. sum(||p(j)|| | j < m) = n
7. j:m, x,y:||p(j)||. x < y (p(j))[x] > (p(j))[y]
8. j:m, x:||p(j)||. (p(j))[x] < n & f((p(j))[x]) = j
9. j: m
10. x:||p(j)||. (p(j))[x] < n & f((p(j))[x]) = j
11. x,y:||p(j)||. x < y (p(j))[x] > (p(j))[y]
12. ||p(j)||1

||p(j)||1

By: InstHyp [0] -3 THEN InstHyp [1] -4 THEN InstHyp [0;1] -4

Generated subgoal:

1 13. (p(j))[0] < n & f((p(j))[0]) = j
14. (p(j))[1] < n & f((p(j))[1]) = j
15. (p(j))[0] > (p(j))[1]
||p(j)||1 5 steps

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(10steps total) PrintForm Definitions Lemmas graph 1 2 Sections Graphs Doc