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There is at most one prime factorization of any natural number (given a range of integers to include).

At: prime factorization unique


  a:{2...}, b:g,h:({a..b}).
  is_prime_factorization(abg)
  
  is_prime_factorization(abh {a..b}(g) = {a..b}(h g = h


By: {Use induction on {a..b}(g), holding a and b constant. }
2 Times Analyze ...w
THEN
(Enough that
(k:g,h:({a..b}).
(is_prime_factorization(abg) & is_prime_factorization(abh)
(
({a..b}(g) = k    {a..b}(g) = {a..b}(h   g = h  {a..b}
(By SideProof)
THEN
CompNatInd Concl ...


Generated subgoal:

1 1. a : {2...}
2. b : 
3. k : 
4. k1:
4. k1<k
4. 
4. (g,h:({a..b}).
4. (is_prime_factorization(abg) & is_prime_factorization(abh)
4. (
4. ({a..b}(g) = k1  {a..b}(g) = {a..b}(h g = h)
5. g : {a..b}
6. h : {a..b}
7. is_prime_factorization(abg)
8. is_prime_factorization(abh)
9. {a..b}(g) = k
10. {a..b}(g) = {a..b}(h)
  g = h

11 steps

About:
intnatural_numberfunctionequalmemberimpliesand
allpfdisp_conclred_hyp
IF YOU CAN SEE THIS go to /sfa/Nuprl/Shared/Xindentation_hack_doc.html

(15steps total) PrintForm Definitions Lemmas FTA Sections DiscrMathExt Doc