```Date: Aug 19, 2013 11:29 AM
Author: Alan Smaill
Subject: Re: Can addition be defined in terms of multiplication?

Alan Smaill <smaill@SPAMinf.ed.ac.uk> writes:> Ben Bacarisse <ben.usenet@bsb.me.uk> writes:>>> William Elliot <marsh@panix.com> writes:>>>>> On Sun, 18 Aug 2013, Peter Percival wrote:> ...>>>> Then I think the onus is on you to produced definitions in one or both of>>>> these forms:>>>>   x + y = ...>>>>   x + y = z <-> ...>>>> >>>> where the only non-logical symbols (baring punctuation) in the>>>> ... are from this set: {*,S,0} or this set: {*,S,0,<}.  I wouldn't>>>> be surprised if + can be defined (in the way requested) from>>>> {*,S,0} or {*,S,0,<} but I would like either to see it spelt out,>>>> or to be given a reference.>>>>>> As Jim Burns said>>> 	z = x + y iff 2^z = 2^x * 2^y>>>>>> where 2^n is defined by induction 2^0 = 1, 2^1 = 1 and 2^(n+1) = 2*2^n>>> all of which can be done with Peano's axioms.>>>> Stepping out of my comfort zone here, but I think the point is that>> allowing recursive definitions makes the theory second-order, and raises>> the question of why one would not simply define + directly that way too.>>>> Broadly speaking, you can either have a second-order theory in which +>> and * and so on are not in the signature of the language (but are>> defined recursively) or you can have a first-order theory where + and *>> and so on are added to the signature, with axioms used to induce the>> usual meaning.>>>> I suspect Peter is talking about a first-order theory where recursive>> definitions are not permitted.>> I do too; it can be done, but it is not easy.  >> See Goedel on defining exponentiation from plus and times via the> Chinese remainder theorem.I should add: while this shows exponentiation is definable (byabbreviational definitions) in Peano Arithmetic, as indeed any primitiverecursive function is, it doesn't show that exponentiation can be sodefined purely in terms of multiplication.-- Alan Smaill
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