Nuprl - Proof: gcd p mul 3 1 1

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At: gcd p mul 3 1 1

1. a :

2. b :

3. y :

4. n :

5. GCD(a;b;y)
6. z :

7. z | n

a
8. z | n

b
9. u :

10. v :

11. GCD(a;b;u

a+v

b)
12. y ~ (u

a+v

z | n

a+v

By:	(Using [`c',u] (FwdThru Thm* a,b,c:. a \| b a \| bc [7])) THEN (Using [`c',v] (FwdThru Thm* a,b,c:. a \| b a \| bc [8]))

Generated subgoal:

1 13. z | nau
14. z | nbv
z | n(ua+vb)
1 step

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IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html