Nuprl - Proof: rem mul 1 1

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At: rem mul 1 1

1. x:
2. y:
3. n:
4. x = (x n)n+(x rem n)
5. |x rem n| < |n|
6. (x rem n) < 0 x < 0
7. (x rem n) > 0 x > 0
8. y = (y n)n+(y rem n)
9. |y rem n| < |n|
10. (y rem n) < 0 y < 0
11. (y rem n) > 0 y > 0
12. (x rem n)(y rem n) = (((x rem n)(y rem n)) n)n+(((x rem n)(y rem n)) rem n)
13. |((x rem n)(y rem n)) rem n| < |n|
14. (((x rem n)(y rem n)) rem n) < 0 (x rem n)(y rem n) < 0
15. (((x rem n)(y rem n)) rem n) > 0 (x rem n)(y rem n) > 0

xy = ((x n)(y n)n+(x n)(y rem n)+(x rem n)(y n)+(((x rem n)(y rem n)) n))n+(((x rem n)(y rem n)) rem n)

By: Subst (xy = ((x n)n+(x rem n))((y n)n+(y rem n))) 0 THEN All ArithSimp

Generated subgoals:

None

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(5steps total) PrintForm Definitions Lemmas graph 1 1 Sections Graphs Doc