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hello welcome to another online lecture for girdles incompleteness theorem I wanted to pick up on something that we didn't quite have enough time to finish or discuss the proof of in class on Monday last Monday so to remind you we were talking about this structure right here this is just the natural numbers with one constant symbol 0 and a function symbol s so the domain of the structure of course is the set of all natural numbers so 0 1 2 3 4 and so on and the interpretation of the constant symbol 0 is going to be the natural number 0 and the interpretation of the this the function symbol s is simply going to be the successor function so in class we talked about this and we're interested really in the theory of this structure but before I talk about that i just want to remind you that we have in this structure we have a way of naming every possible natural number so each natural number corresponds to a sequences of s's followed by a 0 so 0 is 0 s is followed by the constant 0 so that's the constant 0 will be that if you just apply 1 s you get the successor of 0 s squared here or s to the second power is stands for s applied to s apply to 0 and so on so the so the in this way we have a way of naming for every natural number there's a corresponding numeral so this is a term in our language so s to the n 0 this is kind of a macro a way of describing the term which consists of nse's applied to the constant 0 so this is a term in our language that and in this way we every natural number has a specific term in the language now what we're what we're interested in is the language the first order language where the only terms in our language is the constant I'm sorry the only function similar language or the constant symbol s and the constant symbol is the constant symbols 0 and what we're interested in is the theory of our structure natural numbers with successor that's just the set of all sentences so remember a sentence is a is a formula of from this language here that doesn't have any free variables such that that sentence is true in this structure now here's a set of axioms so these are just formulas in our language so the first axiom says that the successor applied to anything is not 0 the second axiom here says that this successor function as a function is one-to-one the third says that every nonzero element must have a predecessor for every why that's not equal to 0 there must exist some X such that Y is the successor of that X and the remaining infinitely many axioms simply say no matter how many times you apply the successor function to some element X you'll never reach X you'll never come back and hit X so you'll never loop back and reach X so a couple of things just sort of obvious things to note if I wanted to so I can prove from this axiom system I can derive for example the fact that two equals two so how do I do that well remember we don't have the way we refer to natural numbers we want to show that the numeral 2 is the same as well the numeral 2 so so I want to somehow show that from this axiom system I can derive the fact that the numeral 2 equals the numeral 2 so what that really means is so I'll write it this way so a s from a s we can derive the fact that the successor of the successor of 0 is equal to the successor of the successor of 0 but this follows because we have just a general principle of first-order logic that if x equals y so if I can prove that x equals y from any axiom system whatever then I can prove that f of X must be equal to F of Y where F is any function symbol actually any term whatsoever so you can always replace equivalence so this is a general law of first-order logic so derive the fact that two equals two all you need is a standard first order logic in this function symbol s what's a little bit more interesting is we need to we would like to be able to also prove that too doesn't equal three so we want to show that the numeral two is not the same cannot be equal to the numeral three well that's actually going to follow directly from s2 and s1 so i'll just give sort of a high-level sketch of what the argument is if i take consider s of s of 0 and s of s of s of three so if those were equal if I could prove that those were equal then I would be able to prove by s2 just take a let x and y let X be s of 0 and Y be a sub s of 0 this allows me to remove one of the one of the s's so at a bateau be able to prove s of 0 is equal to s of s of 0 and again applying s2 again this allows me to prove that 0 is equal to s of 0 but by s1 we know that this is a contradiction because zero can't equal s of anything okay so in this way we see that the ACT the axiom system can be used to prove either that two natural numbers are equal or that two different natural numbers are or the same natural number is equal to itself or the two different natural numbers cannot be equal to each other now it should be I won't go through this proof but the consequences of this axiom system are all true facts about the natural numbers with successor but what we're really interested in showing is that in fact the theory of the natural number successor is just equal to the consequences of that axiom system so that the this theory is in fact oxidizable now we mentioned briefly this is a bit of a digression saying that we're able to axonal ties this theory using this axiom system is not the same thing as saying that the axiom system somehow uniquely identifies or uniquely picks out the structure NS are intended structure so to see that just take any model of this of this axiom system okay so any model must have a domain and a way of interpreting 0 in a way of interpreting s now since we have a way of interpreting 0 we know that we can apply the successor to 0 and I know by i believe it was s2 that applying the six sorry by s1 that applying the successor to 0 I get a new element it can't be 0 itself so i have a new element here and secondly if i apply the successor to this element I get another unique element sorry it's a successor of the successor of 0 and I can keep doing that so applying successors as many times as I want and what I get is of course in the structure I have a copy so formally it's an isomorphic copy of this structure here but there might be some element a in the domain such that a is different from 0 or the successor of 0 or the successor of the successor of 0 and so on so I might have some element in the domain that is what you might call non-standard a non-standard element of the domain and and all you mean by non-standard is it's just not a natural number now if i have this element a then i know that i can apply the successor to it the successor function to it and i know that this has to be different from a and furthermore again the same argument opie reply so i can keep applying successor to a to get an infinite sequence of natural numbers going in the right direction but since i know a is different from 0 then a has to have a predecessor so there has to be an element star such that when i applied the successor to star i get a but star itself cannot be 0 and send star itself because if star itself with 0 then a must be s of 0 but we know that the successor function has to be has to be a one to one and so sorry can't actually be 0 because we are supposing that a is not equal to s of 0 for any of the four is not equal to any natural number so star itself can't be 0 that means that star has to have its self a predecessor let's call it star Prime and the same argument applies and so we actually have a copy of the integers Z so this turns out to be a copy of the integers of Z so we can actually characterize what all the structures of all the structures which which make all of the Act all the consequences of the axiom is true in the following way well say that two elements of our domain are equivalent if we can apply s a finite number of times to a in order to yield be and it's easy to see that this is an equivalence relation on a and so we have one of the equivalence classes must be an isomorphic copy of our intended structure but they're going to be all these other equivalence classes which each one has to be isomorphic copies of Z now we might have countably many equivalence classes all together or uncountably many equivalence classes all together but the important point is that just because we we know and this is what will we're about to prove that we can axia ties this structure using these axioms the axioms are still not strong enough to uniquely pin down our intended structure all right now in order to prove that are so let me go back here we want to show remember what we're trying to show is that this theory is axion tized by by this set of axioms actually I'm not going to prove that I'm going to prove something a little bit stronger namely that the theory itself this theory and s is decidable so there's an algorithm that decides for any formula whether or not it is a true in the structure or not so there's an algorithm that decides this set and because if the theory is decidable then it has to be recursively enumerable or effectively innumerable and we saw in class that every effectively innumerable set of sentences has to be accident I zabal now this doesn't show that this set of sentences AAS in particular AXA meant eise's this structure that requires a separate argument and and I'll leave it up to you to work out those details so what we really want to prove is that the theory T the theory of natural numbers of successor is in fact decidable and the way we're going to prove that is we're going to show that it admits elimination of quantifiers now what that means is that for any formula Phi and this formula Phi might have lots and lots of quantifiers in it I can always find a quantifier free formula sigh such that my theory t thinks that Phi is equivalent to sigh so in order to determine whether or not Phi follows from from T so whether or not Phi is a consequence of the theory T it's enough to check this formula sigh this quantifier free formula sigh now why does this give us decidability well given an arbitrary formula Phi I can in a decidable way or an algorithmic way find the formulas I that is equivalent to Phi so in order to determine whether or not Phi is an element of my theory or follows from my theory it's enough to check whether or not sigh follows from my theory now proving a limitation of quantifiers is not always so easy but fortunately fortunately there's a theorem that makes our job a little bit easier all we really have to do is check not that every formula Phi is equivalent to a quantifier free formula but rather we simply have to show that formulas of a very special form namely forms of the form there exists in X such that a finite conjunction of formulas each of which is an atomic formula or the negation of a Tom formula is quantifier free so in other words I can find a quantifier a quantifier free formula sigh that is equivalent to this formula which is of a very special form I'll sort of just State this theorem without proof but the proof follows by induction on the structure of Phi so the way it goes is you suppose that for every formula of this form I can find a quantifier free formula size such that T thinks that faience I are equivalent then I can prove that in fact I can lift that to all possible formulas in my language ok all right so now let's show that this theory admits the elimination of quantifiers by the previous theorem it is enough so it is enough to consider formulas of the form there exists X alpha 0 and and alpha and so I need to consider formulas only of this form right here now each alpha I is either going to be an atomic formula or the negation of atomic formula so an atomic formula will be something of the form s to the n of let's say let me write it as s to the end of V equals s to the n of you or s to the n sorry s to the M of you or s to the n V is not equal to s to the M of you where v and u are variables or 0 so each alpha I is an equation or the negation of any operation now it's easy to see we can just assume that V that X must occur in each of the equations so we assume X must occur in each alpha I the reason is if X doesn't occur in alpha I then let's just take so there exists X alpha and beta if X doesn't occur in alpha here this is just logically equivalent to alpha and there exists X beta so we can pull out all the Khan junks that that in which X doesn't actually occur so we can assume that X occurs in each of those alpha eyes so each alpha I is of the form s to the n x equals s to the MU or s to the n X does not equal s to the M you now by the observation i made earlier we also can assume that you is in fact different from X now why is that well if u is equal to X then I can just simply replace s to the n x equals s to the MU with so if n equals M I can replace that conjunct with 0 equals 0 and if n is not equal to M i can replace it with the conjunct with 0 doesn't equal 0 so if u is equal to X I can just by looking at how many s's there are I can either say that definitely has to be true or that definitely is not true so I can replace that conjunct with a simpler formula 0 equals 0 or 0 doesn't equal zero and these don't of course have X in it so by the previous argument I can pull those out okay so now we're in a situation where we have this formula there exists X alpha 0 and and alpha N and each alpha I is of the form s to the n x equals s to the M you or the negation of that as CN X does not equal s to the M you now we have two cases case one is so either all of the Alpha eyes are going to be the negation of the equation so alpha so each alpha I is of the form s to the n X does not equal s to the M you so all of the alpha i's are the negation of an equation or there's at least one that is not the negation of an equation that will be case too but for now let's just consider case one if that's the case then I claim we can replace the whole formula there exists X alpha 0 and and alpha N with zero equals 0 that's just simply de Rivel so the intuition here is that if it's the negation of all of these equations right here I just have to that that's easy to see that we can derive the negation of well so the intuition is that it's very easy to make these negations true because we can just set X to be so suppose u is equal to 0 in this case and just let X be anything so that n is different from m okay so if all of the Alpha eyes are negations of equations i can replace this entire formula just simply with the formula 0 equals 0 this is I didn't really give you the argument I give you a high-level sketch of the argument but let's just this should be a little a moment's reflection should tell you that this is true but what we're really interested in is the second case so this is case 2 and in case 2 there is some alpha I such that alpha i is equal to s to the n x equals s to the M you okay so sdn x equals s to the M you now remember that you here does not contain X now I claim that i can simply replace this conjunct with let's call this the term t so i'll say that s to the MU is the term t i can replace this conjunct so we can replace alpha I with T is not equal to 0 and T is not equal to s of 0 and T is not equal to s to the n minus 1 of 0 because we can because this conduct whether or not this contract is well that we can just talk about true rather than drivable in the axiom system whether or not this is true depends simply on whether or not the s's there are NSS here matches the number of s's over here and of course if n is equal to if n is 20 if n is equal to zero then replace alpha I with zero equals zero now once we've done that so we replace alpha I with this equation T naught equal to or this conjunction T naught equal to 0 and T naught equal to 1 and T naught equal to dot dot we look at all of the other formulas that we see so we look at all of the other con junks that we see so we have remember we have there exists X alpha 0 and and then we have this alpha I which we're dealing with and alpha n now suppose for example there's another alpha J which is of the form s to the K X is equal to for example you where you is some term well we shouldn't use use let's say V where V is some term now what we do is we simply say so we add esas to both sides so that we have a so we we say we note that if this is equal then it's the case that s to the k plus M X is equal to s to the m/v so we add esas to both sides of this equation at exactly sorry it shouldn't be M it should be n esas to both sides now once we do that we can turn this equation remembering that s to the n X is actually just equal to t2 this term t we can turn this into the equation s to the K T is equal to s to the n V where V is so term in our language so I'm able to take any alpha J which is of this form I add s is to it and I turn it into an equation of this form s to the K T equals s to the n V so I replace alpha J with this equation right here so if this is true then this equation has to also be true as well if the same would hold if we have some other alpha J prime which is equal to s to the K X is not equal to V the exact same argument hold except we replace this with s2 the katie is not equal to s to the n V so what I've been able to do is I take my formula there exists alpha 0 and and alpha I and and alpha N and I replace alpha I with this equation t is not equal to 0 and t is not equal to s of 0 and dot dot and t is not equal to let's let's say s to the n minus 1 of 0 then I take all of these other formulas and I replace them with equations of the form it will essentially be s to the K T equals s to the n V that's re s to the n V and that's true for all of these other formula so what I've been able to do is take all of these contexts and remove the exes so note that X doesn't appear X doesn't appear in any conjunct so we can drop the quantifier so I've been able to take each of these conjunct remove all of the exes from the conjunct so that makes this quantifier useless so we drop we go ahead and drop that quantifier so I'm able to turn any formula of this form into a formula that doesn't include a quantifier and that actually compete completes the proof given that are given our earlier theorem so we've been able to show that the theory of the NS admits elimination of quantifiers a little bit of extra argument gives you that that tells you that the theory is decidable and so it must be ax minute I zabal in order to show that it's accident I zabal by the set of axioms we discussed earlier in this lecture that requires a little bit of extra proof essentially what you want to show is all the argument I just cave proving that the theory of NS admits elimination of quantifiers can be done using just the axioms a s