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	<title>Comments on: Re: Just what *is* Arithmetic with multiplication only?</title>
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		<title>By: admin</title>
		<link>http://www.aboutlogic.info/re-just-what-is-arithmetic-with-multiplication-only/comment-page-1#comment-5565</link>
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		<pubDate>Wed, 10 Mar 2010 02:01:09 +0000</pubDate>
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  &lt;p&gt;Bill Taylor &lt;math...@math.canterbury.ac.nz&gt; a écrit dans le message : &lt;br /&gt; 89n3uc$34...@cantuc.canterbury.ac.nz... &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&lt;/p&gt;&lt;p&gt;- Hide quoted text -- Show quoted text -&lt;/p&gt;&gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with &lt;br /&gt; &gt; addition (and successor) but NOT multiplication are decidable, are often &lt;br /&gt; &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; &lt;br /&gt; multiplication &lt;br /&gt; &gt; or addition. &lt;br /&gt; &lt;p&gt;&gt; Now I know of the standard way that addition can be *defined* in terms of &lt;br /&gt; &gt; multiplication and successor, but I&#039;m not sure that this has too much to &lt;br /&gt; do &lt;br /&gt; &gt; with the oft-suggested implication that &quot;arithmetic with multiplication &lt;br /&gt; only &lt;br /&gt; &gt; is decidable&quot;. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Is that last statement ever made, explicitly? &#160;I don&#039;t recall it being so. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; And just what would it BE - arithmetic with multiplication and successor, &lt;br /&gt; &gt; but not addition. &#160;What sensible axioms could it be given? &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Help please. &lt;br /&gt; &lt;br /&gt;A good reference for decidability of Th(N;=,x) (the elementary theory of &lt;br /&gt; integer multiplication - sometimes called &quot;Skolem Arithmetic&quot;) is : &lt;br /&gt; C.Smorynski, &quot;Logical Number Theory I&quot;, Springer-Verlag, &#160;1991. &lt;br /&gt; It presents Cegielski&#039;s proof (1981) by quantifier elimination. &lt;br /&gt; Alternative proofs were given by Mostowski (&quot;On directs products of &lt;br /&gt; theories&quot;,J.Symb.Log. 17, 1-31, 1952) and later Hodgson (&quot;On direct products &lt;br /&gt; of automaton decidable theories&quot;, Theor.Comp.Sci. 19,331-335, 1982) -- both &lt;br /&gt; use Presburger&#039;s decidability result for Th(N;=,+). &lt;br /&gt; Alexis Bes &lt;br /&gt;
  
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		<content:encoded><![CDATA[<p>Bill Taylor &lt;math&#8230;@math.canterbury.ac.nz&gt; a écrit dans le message : <br /> 89n3uc$34&#8230;@cantuc.canterbury.ac.nz&#8230;  </p>
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<p>&gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with <br /> &gt; addition (and successor) but NOT multiplication are decidable, are often <br /> &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; <br /> multiplication <br /> &gt; or addition. <br /> 
<p>&gt; Now I know of the standard way that addition can be *defined* in terms of <br /> &gt; multiplication and successor, but I&#8217;m not sure that this has too much to <br /> do <br /> &gt; with the oft-suggested implication that &quot;arithmetic with multiplication <br /> only <br /> &gt; is decidable&quot;.  </p>
<p>&gt; Is that last statement ever made, explicitly? &nbsp;I don&#8217;t recall it being so.  </p>
<p>&gt; And just what would it BE &#8211; arithmetic with multiplication and successor, <br /> &gt; but not addition. &nbsp;What sensible axioms could it be given?  </p>
<p>&gt; Help please. </p>
<p>A good reference for decidability of Th(N;=,x) (the elementary theory of <br /> integer multiplication &#8211; sometimes called &quot;Skolem Arithmetic&quot;) is : <br /> C.Smorynski, &quot;Logical Number Theory I&quot;, Springer-Verlag, &nbsp;1991. <br /> It presents Cegielski&#8217;s proof (1981) by quantifier elimination. <br /> Alternative proofs were given by Mostowski (&quot;On directs products of <br /> theories&quot;,J.Symb.Log. 17, 1-31, 1952) and later Hodgson (&quot;On direct products <br /> of automaton decidable theories&quot;, Theor.Comp.Sci. 19,331-335, 1982) &#8212; both <br /> use Presburger&#8217;s decidability result for Th(N;=,+). <br /> Alexis Bes </p>
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		<title>By: admin</title>
		<link>http://www.aboutlogic.info/re-just-what-is-arithmetic-with-multiplication-only/comment-page-1#comment-5564</link>
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		<pubDate>Wed, 10 Mar 2010 02:01:08 +0000</pubDate>
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  &lt;p&gt;&lt;/p&gt;&lt;p&gt;- Hide quoted text -- Show quoted text -&lt;/p&gt;Bill Taylor wrote: &lt;br /&gt; &gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with &lt;br /&gt; &gt; addition (and successor) but NOT multiplication are decidable, are often &lt;br /&gt; &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication &lt;br /&gt; &gt; or addition. &lt;br /&gt; &lt;p&gt;&gt; Now I know of the standard way that addition can be *defined* in terms of &lt;br /&gt; &gt; multiplication and successor, but I&#039;m not sure that this has too much to do &lt;br /&gt; &gt; with the oft-suggested implication that &quot;arithmetic with multiplication only &lt;br /&gt; &gt; is decidable&quot;. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Is that last statement ever made, explicitly? &#160;I don&#039;t recall it being so. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; And just what would it BE - arithmetic with multiplication and successor, &lt;br /&gt; &gt; but not addition. &#160;What sensible axioms could it be given? &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Help please. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; ------------------------------------------------------------------------------- &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160;Bill Taylor &#160; &#160; &#160; &#160; &#160; &#160;W.Tay...@math.canterbury.ac.nz &lt;br /&gt; &gt; ------------------------------------------------------------------------------- &lt;br /&gt; &gt; Godel&#039;s humility theorem:- &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; Providing you are capable of doing sufficient math, &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;you are intelligent if and only if you cannot prove you are. &lt;br /&gt; &gt; ------------------------------------------------------------------------------- &lt;br /&gt; &lt;br /&gt;&#160; &#160; Goedel&#039;s complementary humility theorem: &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;Providing you are capable doing sufficient physics, &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;you are intelligent if and only if you can prove you &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;are ignorant of Goedel&#039;s theorem. &lt;br /&gt;
  
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<p>Bill Taylor wrote: <br /> &gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with <br /> &gt; addition (and successor) but NOT multiplication are decidable, are often <br /> &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication <br /> &gt; or addition. <br /> 
<p>&gt; Now I know of the standard way that addition can be *defined* in terms of <br /> &gt; multiplication and successor, but I&#8217;m not sure that this has too much to do <br /> &gt; with the oft-suggested implication that &quot;arithmetic with multiplication only <br /> &gt; is decidable&quot;.  </p>
<p>&gt; Is that last statement ever made, explicitly? &nbsp;I don&#8217;t recall it being so.  </p>
<p>&gt; And just what would it BE &#8211; arithmetic with multiplication and successor, <br /> &gt; but not addition. &nbsp;What sensible axioms could it be given?  </p>
<p>&gt; Help please.  </p>
<p>&gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Bill Taylor &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;W.Tay&#8230;@math.canterbury.ac.nz <br /> &gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> &gt; Godel&#8217;s humility theorem:- <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Providing you are capable of doing sufficient math, <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;you are intelligent if and only if you cannot prove you are. <br /> &gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- </p>
<p>&nbsp; &nbsp; Goedel&#8217;s complementary humility theorem: <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Providing you are capable doing sufficient physics, <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;you are intelligent if and only if you can prove you <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;are ignorant of Goedel&#8217;s theorem. </p>
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		<title>By: admin</title>
		<link>http://www.aboutlogic.info/re-just-what-is-arithmetic-with-multiplication-only/comment-page-1#comment-5563</link>
		<dc:creator>admin</dc:creator>
		<pubDate>Wed, 10 Mar 2010 02:01:07 +0000</pubDate>
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  &lt;p&gt;&lt;/p&gt;&lt;p&gt;- Hide quoted text -- Show quoted text -&lt;/p&gt;Bill Taylor wrote: &lt;br /&gt; &lt;p&gt;&gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with &lt;br /&gt; &gt; addition (and successor) but NOT multiplication are decidable, are often &lt;br /&gt; &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication &lt;br /&gt; &gt; or addition. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Now I know of the standard way that addition can be *defined* in terms of &lt;br /&gt; &gt; multiplication and successor, but I&#039;m not sure that this has too much to do &lt;br /&gt; &gt; with the oft-suggested implication that &quot;arithmetic with multiplication only &lt;br /&gt; &gt; is decidable&quot;. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; Is that last statement ever made, explicitly? &#160;I don&#039;t recall it being so. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; And just what would it BE - arithmetic with multiplication and successor, &lt;br /&gt; &gt; but not addition. &#160;What sensible axioms could it be given? &lt;br /&gt; &lt;br /&gt;Consider--add to suitable* axioms for first order logic these: &lt;br /&gt; 1. if (x = y) then (if (x = z) then (y = z)) &lt;br /&gt; 2. if (x = y) then (sx = sy) &lt;br /&gt; 3. 0 &lt;&gt; sx &lt;br /&gt; 4. if (sx = sy) then x = y &lt;br /&gt; 5. x + 0 = x &lt;br /&gt; 6. x + sy = s(x + y) &lt;br /&gt; 9A. for any wff A, if A0 then (if (all x)(if Ax then Asx) then (all &lt;br /&gt; x)Ax) &lt;br /&gt; Then you have Presburger&#039;s 1929 axioms. &lt;br /&gt; Now replace those with: &lt;br /&gt; 1. if (x = y) then (if (x = z) then (y = z) &lt;br /&gt; 2. if (x = y) then (sx = sy) &lt;br /&gt; 3. 0 &lt;&gt; sx &lt;br /&gt; 4. if (sx = sy) then x = y &lt;br /&gt; 7. x.0 = 0 &lt;br /&gt; 8n. for each natural number n, N0.(sy) = N(N0.y), where &quot;N&quot; is n &#039;s&#039;s. &lt;br /&gt; 9A. for any wff A, if A0 then (if (all x)(if Ax then Asx) then (all &lt;br /&gt; x)Ax) &lt;br /&gt; and you have _my_ candidate for arithmetic without +. &lt;br /&gt; 8n is just an axiom schema to replace &lt;br /&gt; 8. x.(sy) = (x.y) + x &lt;br /&gt; by replacing &quot;+ x&quot; with x-many successor function applications. &lt;br /&gt; Could be rubbish--no guarantee of fitness for purpose! &lt;br /&gt; *Note, 1. is needed because the &quot;suitable&quot; fol has no axioms for &quot;=&quot;, it &lt;br /&gt; is introduced as a binary predicate. &lt;br /&gt; All of these &quot;s&quot;s remind me to remind you that &quot;Presburger&quot; only has &lt;br /&gt; one. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;Regards, PP &lt;br /&gt; &lt;/p&gt;&lt;p&gt;- Hide quoted text -- Show quoted text -&lt;/p&gt;&lt;p&gt;&gt; Help please. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;&gt; ------------------------------------------------------------------------------- &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160;Bill Taylor &#160; &#160; &#160; &#160; &#160; &#160;W.Tay...@math.canterbury.ac.nz &lt;br /&gt; &gt; ------------------------------------------------------------------------------- &lt;br /&gt; &gt; Godel&#039;s humility theorem:- &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; Providing you are capable of doing sufficient math, &lt;br /&gt; &gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;you are intelligent if and only if you cannot prove you are. &lt;br /&gt; &gt; ------------------------------------------------------------------------------- &lt;br /&gt; &lt;br /&gt;
  
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<p>Bill Taylor wrote: <br /> 
<p>&gt; Accounts of Pressburger Arithmetic, and the proof that arithmetic with <br /> &gt; addition (and successor) but NOT multiplication are decidable, are often <br /> &gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication <br /> &gt; or addition.  </p>
<p>&gt; Now I know of the standard way that addition can be *defined* in terms of <br /> &gt; multiplication and successor, but I&#8217;m not sure that this has too much to do <br /> &gt; with the oft-suggested implication that &quot;arithmetic with multiplication only <br /> &gt; is decidable&quot;.  </p>
<p>&gt; Is that last statement ever made, explicitly? &nbsp;I don&#8217;t recall it being so.  </p>
<p>&gt; And just what would it BE &#8211; arithmetic with multiplication and successor, <br /> &gt; but not addition. &nbsp;What sensible axioms could it be given? </p>
<p>Consider&#8211;add to suitable* axioms for first order logic these: <br /> 1. if (x = y) then (if (x = z) then (y = z)) <br /> 2. if (x = y) then (sx = sy) <br /> 3. 0 &lt;&gt; sx <br /> 4. if (sx = sy) then x = y <br /> 5. x + 0 = x <br /> 6. x + sy = s(x + y) <br /> 9A. for any wff A, if A0 then (if (all x)(if Ax then Asx) then (all <br /> x)Ax) <br /> Then you have Presburger&#8217;s 1929 axioms. <br /> Now replace those with: <br /> 1. if (x = y) then (if (x = z) then (y = z) <br /> 2. if (x = y) then (sx = sy) <br /> 3. 0 &lt;&gt; sx <br /> 4. if (sx = sy) then x = y <br /> 7. x.0 = 0 <br /> 8n. for each natural number n, N0.(sy) = N(N0.y), where &quot;N&quot; is n &#8216;s&#8217;s. <br /> 9A. for any wff A, if A0 then (if (all x)(if Ax then Asx) then (all <br /> x)Ax) <br /> and you have _my_ candidate for arithmetic without +. <br /> 8n is just an axiom schema to replace <br /> 8. x.(sy) = (x.y) + x <br /> by replacing &quot;+ x&quot; with x-many successor function applications. <br /> Could be rubbish&#8211;no guarantee of fitness for purpose! <br /> *Note, 1. is needed because the &quot;suitable&quot; fol has no axioms for &quot;=&quot;, it <br /> is introduced as a binary predicate. <br /> All of these &quot;s&quot;s remind me to remind you that &quot;Presburger&quot; only has <br /> one.  </p>
<p>Regards, PP  </p>
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<p>&gt; Help please.  </p>
<p>&gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Bill Taylor &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;W.Tay&#8230;@math.canterbury.ac.nz <br /> &gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> &gt; Godel&#8217;s humility theorem:- <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Providing you are capable of doing sufficient math, <br /> &gt; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;you are intelligent if and only if you cannot prove you are. <br /> &gt; &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- </p>
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		<title>By: admin</title>
		<link>http://www.aboutlogic.info/re-just-what-is-arithmetic-with-multiplication-only/comment-page-1#comment-5562</link>
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		<pubDate>Wed, 10 Mar 2010 02:01:06 +0000</pubDate>
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  Accounts of Pressburger Arithmetic, and the proof that arithmetic with &lt;br /&gt; addition (and successor) but NOT multiplication are decidable, are often &lt;br /&gt; preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication &lt;br /&gt; or addition. &lt;br /&gt; &lt;p&gt;Now I know of the standard way that addition can be *defined* in terms of &lt;br /&gt; multiplication and successor, but I&#039;m not sure that this has too much to do &lt;br /&gt; with the oft-suggested implication that &quot;arithmetic with multiplication only &lt;br /&gt; is decidable&quot;. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;Is that last statement ever made, explicitly? &#160;I don&#039;t recall it being so. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;And just what would it BE - arithmetic with multiplication and successor, &lt;br /&gt; but not addition. &#160;What sensible axioms could it be given? &lt;br /&gt; &lt;/p&gt;&lt;p&gt;Help please. &lt;br /&gt; &lt;/p&gt;&lt;p&gt;------------------------------------------------------------------------------- &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160;Bill Taylor &#160; &#160; &#160; &#160; &#160; &#160;W.Tay...@math.canterbury.ac.nz &lt;br /&gt; ------------------------------------------------------------------------------- &lt;br /&gt; Godel&#039;s humility theorem:- &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; Providing you are capable of doing sufficient math, &lt;br /&gt; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160;you are intelligent if and only if you cannot prove you are. &lt;br /&gt; ------------------------------------------------------------------------------- &lt;br /&gt;
  
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		<content:encoded><![CDATA[<p>Accounts of Pressburger Arithmetic, and the proof that arithmetic with <br /> addition (and successor) but NOT multiplication are decidable, are often <br /> preceded by a comment or so about arithmetic with &quot;only one of&quot; multiplication <br /> or addition. <br /> 
<p>Now I know of the standard way that addition can be *defined* in terms of <br /> multiplication and successor, but I&#8217;m not sure that this has too much to do <br /> with the oft-suggested implication that &quot;arithmetic with multiplication only <br /> is decidable&quot;.  </p>
<p>Is that last statement ever made, explicitly? &nbsp;I don&#8217;t recall it being so.  </p>
<p>And just what would it BE &#8211; arithmetic with multiplication and successor, <br /> but not addition. &nbsp;What sensible axioms could it be given?  </p>
<p>Help please.  </p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Bill Taylor &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;W.Tay&#8230;@math.canterbury.ac.nz <br /> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- <br /> Godel&#8217;s humility theorem:- <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Providing you are capable of doing sufficient math, <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;you are intelligent if and only if you cannot prove you are. <br /> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- </p>
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