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June 23rd, 2014, 10:56 AM  #1 
Newbie Joined: Jun 2014 From: here Posts: 2 Thanks: 0  H&B's 2nd axiom for successor function contradicts 1st?
Hi all, I'm reading Charles Petzold's "The Annotated Turing" for fun, and I'm no mathematician, so this might be a silly question. I have a question about page 226, where Hilbert and Bernay's axioms for the successor function are defined. Specifically, the second axiom: (Ǝx)(y)S(y,x) I don't agree with this axiom: what's an example of a number with no successor? Doesn't this directly contradict the first axiom: (x)(Ǝy)S(x,y) ? Thanks. 
June 23rd, 2014, 01:12 PM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
For all x, there is some y such that y is the successor of x. Not the same as the above because of the order of quantifiers and the order of the arguments. 0 may have no predecessor but it and all other elements have successors.  
June 24th, 2014, 03:46 AM  #3 
Newbie Joined: Jun 2014 From: here Posts: 2 Thanks: 0 
Thanks. I didn't know these only applied to the natural numbers, so yes, they both make sense in that regard.

June 24th, 2014, 05:43 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  

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1st, 2nd, axiom, contradicts, function, handb, question, successor 
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