 Welcome back in the last class under section axiomatic propositional logic we presented Russell Whitehead's axiomatic system that is what we find it in the book principia mathematical in one of the sections on deductions you will find some interesting proofs such as law of identity law of non contradiction and many other important theorems so what is our main goal may our main goal was this that all the valid formulas in your formal axiomatic system should find a proof so that is the reason why we are doing this axiomatic propositional calculus so today we will be presenting another axiomatic system which is due to another two sets set of mathematicians Hilbert and Ackerman Hilbert is famous for his different challenges and all Hilbert they are called as Hilbert problems so we will not be talking about all the problems and all but so we are presenting a kind of axiomatic system which is proposed by Hilbert and Ackerman so what Hilbert's interest was mainly axiomatizing geometry so he lived from 1862 to 1943 his dream was indeed to create a perfect axiomatic system for mathematics or at least if you restrict ourselves to at least geometry in arithmetic as far as geometry in arithmetic are concerned he wants to come up with a grand axiomatic system if you can construct this axiomatic system then what will happen is this that all possible true statements can be proved or you can show that all the provable theorems can be are automatically true so now according to him the consistency is a part of mathematics such as in the case of natural numbers it was established by some kind of finitary methods which could not lead to any contradiction to start with tautologies you will never end up with a contradiction then this part can be used as a secure foundations for the entire mathematics so you showed that by using some kind of finitary methods you could come up with some proofs which are devoid of contradictions and that can be used as a pillar for constructing some other kind of theorems so it is only till Godel Godel has come up with an interesting theorem which is called as incomprehendence theorem according to which for any formal axiomatic system as is the case of a principle of mathematics are the Hilbert Ackermann axiomatic system so he has come up with a radical kind of view that is this that there is there are always some statements about natural numbers that is in the arithmetic which are obviously considered to be true which cannot be proven within the system with using its own axioms and with using its own principles etc. So that means system leads to incomprehendence what is incomprehendence anything which is probable is true or if you can show it to be valid and all the valid formulas have to be have to find a proof if that is the case then your system is considered to be complete so now Godel has come up with an interesting kind of theorem with which you one can know one can show that no consistency system can be used to prove its own consistency if you think that you know principia Mathematica and Hilbert Ackermann system they are considered to be consistent anyway Hilbert's dream has been shattered by this one of the important theorems in logic that is Godel's incomprehendence theorem which we will talk about it under the limitations of the first order logic. So as far as prepositional logics are concerned they are all decidable and complete and consistent and sound but whereas in the case of first order logics they are semi decidable and at least to incomprehendence etc. So in the context of first order logic when I talk about predicate logic these things will become prominent. So our goal is to present in this lecture is to present Hilbert Ackermann axiomatic system and then we will be proving some important theorems such as P ? P or ? P etc and then not only that thing we will be making use of one important theorem prepositional logic axiomatic prepositional logic. So that is the deduction theorem if you can use deduction theorem with the set of axioms that you already have then our proofs will become simpler. So now to start with Hilbert Ackermann axiomatic system it is presented in various ways in various textbooks in particular in most of the test books these are the three axioms that are provided in the textbooks they like Mendelssohn etc. So these are some of the axioms you have to note that Hilbert Ackermann makes use of only two primitive logical symbols that is implication and negation. So these are only two symbols which you commonly find it in you find it in the Hilbert Ackermann axioms. So to start with any axiomatic system consist of a set of axioms and transformation rules and of course the rule of detachment. So instead of five axioms as is the case of Bertrand Russell invited we have only three axioms here and all these axioms are in are expressed in terms of implication and negation. The first axiom is A ? B ? A and the second one A ? B ? C ? A ? B ? A ? C and now this third axiom first stated like this in the beginning not B ? 0 A is A ? B but in some textbooks like Mendelssohn introduction to mathematical logic you will find this particular kind of thing not B ? 0 A ? 0 B ? A ? B but when you can show that if you take only the first three axioms that will be that will constitute a formal axiomatic system and then if you take the first two axioms one and two and the fourth one which is there the top of the slide then you can formulate a different kind of axiomatic system let us say H ? So now we can easily show that this H and H ? are more or less they are same if you can somehow show that this ? B ? 0 A ? B will get it as an outcome of this revised kind of axiom that is ? B ? 0 A ? B then you can show that these two axiomatic systems are similar to each other. So I will be taking into consideration the first two axioms and the revised axiom which was presented by Mendelssohn in the tradition of Hilbert Ackermann axiomatic system. So now using this axioms what essentially we are trying to do is we are trying to prove some important theorems. So now so far we have seen some important theorems such as P ? B in Russell whitehead axiomatic system. So now we try to show with the help of this Hilbert Ackermann axiomatic system you will be proving some of the important theorems such as A ? A now any axiomatic system this law of identity should come as an outcome. So now these are the important things that we need to know these are the axioms you can write HA1 H stands for Hilbert Ackermann and then this is the axiom number 1 A ? B ? A HA2 A ? B ? C ? A ? B ? A ? C so our third one is this ? B ? 0 A this is what we are going to take into consideration the revised axiom this means ? B ? A ? B this brackets needs to be closed properly and you want you can write something and the rule of detachment is as it is suppose if we can assume that A and if we can assume that A ? B from this these two things you will get B as an outcome. So we will be making use of these things on the right hand side of the board improving some of the important theorems then we later we will use another one which is called as deduction theorem so which we will employ it little bit later. So now we are trying to show A ? A by using only these axioms and some kind of transformation rules plus this detachment rule so now what exactly we are trying to do is we are trimming these axioms in such a way that it leads to another kind of truth these are all obviously true statements you trim it in such a way so that it transforms into this particular kind of preposition. So now for this you start with the Hilbert Ackermann axiomatic system to that is A ? B ? C same thing which we are writing it again A ? B ? A ? C so now in this what you will do is for wherever you find B you substitute it as A ? E and wherever you find C you substitute with A and of course even B also as E so that means you are replacing B with A and you are replacing sorry you are replacing B with A ? A C with A and this is not required so these are the two operations that you are trying to do so the rational behind this thing is that if you substitute anything into this axiom uniformly that will retain is tautology hood that means it is still it will still act like a tautology it is a total so now this is the first step and the second step is a is as it is because we are proving A ? A that means you have to eliminate this B sense is somehow so that you will find only a senior formula so that is the reason why we use this uniform substitution rule so now what is B here B means A ? A and then C is also considered to be a now the second one A ? A ? A B is A ? so that is the first one and then second one is A ? A so ultimately we ensured that the last step of your condition the consequent of your conditional at the occupying the last position is somewhat closer to what we are trying to prove so that means A ? A so now what one needs to do is somehow we need to detach the whole thing and somehow you get this particular kind of thing which that is what we design so now this is step number 2 so now we have an axiom A ? B ? A so now in this you substitute again for B you substitute A ? C so now this becomes what is this this is axiom number 1 HA 1 if we can understand one proof then we can solve we can prove many other theorems you know so this will become instead of B we write A ? A and then this is as it is so this is what instance of Hilbert-Akerman axiomatic system 1 that means you substituted A ? A for B and this is what you get so now the fifth step you observe these two things so this is same as this one so now these two what are these things two and four you have to write justification here modus ponens that means we use this kind of rule then this gets detached and then whatever remains is this one A ? A ? A is so that is A ? A ? A this part goes enough it gets detached and whatever remains is this portion whatever is there afterwards implies A ? A so this is how did we get this one by applying modus ponens on two and four so now in the sixth step still it is not in this particular kind of form somehow we need to detach this thing so how do we detach this particular kind of thing somehow we need to again fall back on our axioms if you can use any one of these two things you will not get to this particular kind of form but if you use this one you transform it in such a way that for example if instead of B you put A here then it will become A ? A ? A so now A ? A ? A what is this this is an instance of axiom number 1 because instead of B you have put A here uniformly is substituted A for B so this is what you get so now under the eighth step so these two are same A ? A ? A and A ? A ? A this is like X and X ? Y so we will get Y so now in the eighth step you will get A so now what is that we are seeing in this particular kind of proof you might come up with A ? A in maybe in less number of steps but this appears to be the case that at least seven or eight steps are involved in proving A ? A so there are at least two things which need to note I start with the axioms and then you I trim these axioms in such a way that I will form this particular kind of theorems so these theorems might come in four steps sometimes seven steps or sometimes if your axiom the choice of your axioms are wrong then you will be you will be playing with it and ultimately it might it might take some 16 steps sometimes proofs might take even days also so the effective proof is considered to be that particular kind of proof which ends in fine steps in finite intervals of time that means if a proof never ends in it goes on and on and on and that is not considered to be an effective kind of proof so you should also note that all these proofs you can transform it into some kind of language programming language and then you can talk about this particular kind of thing or you can develop a software in which to give the feedback of all these axioms in all and then you put this data A ? A whether it is a theorem or not that software will tell us whether this is a theorem or there are some software which provide even proofs also that is not what we are going into the details of it so as a first step we are trying to show how we can generate aim place here by producing some kind of rigorous proof so with this proof what what else we can find out is this that everything is listed here that means everything is stated explicitly in terms of axioms which are considered to be obviously true and then the modus ponens rule which is a truth preserving rule and transformation rules also preserves the truth so everywhere we are everywhere the you considered as hypotheses or premises they are all true and the final step of your proof is called usually called as a theorem that is what we are said in the beginning of discussing this particular kind of axiomatic propositional logic so this is the first theorem which you will get it as an outcome and from this by definition you can say it is A or A so this is also can be proved if this is proved then not A or A can also come as a theorem so now what we will be doing is so we will be proving some other kinds of theorems such as not not be lies be so let us guess ourselves familiarize with this particular kind of axiomatic system so what essentially we are trying to do is that so we have formulated an axiomatic system then in that axiomatic system which consist of only few rules and using this only if you have only few axioms and very minimal set of rules and all and with that you generate all kinds of true statements that are theorems so you are not supposed to use anything outside these three axioms so if you use anything outside the things and all like in the case of Euclidean axiomatic system that is also considered to be a formal axiomatic system but the problem there was is that the proofs are not rigorous like these proofs and all in the sense that there are many implicit assumptions which are part and parcel of your proof and there are certain things which are not part of the proof also take they also took part in the proof so in that sense Euclidean axiomatic system is not so rigorous like the axiomatic system that we are trying to present the very purpose of presenting this axiomatic system is to get rid of those non rigorous kind of proofs so in this thing everything is stated explicitly there is nothing hidden no hidden assumptions are there so everything comes through by trimming these axioms you will get your theorems so now let us try to prove another theorem which we have already proved it in the in the axiomatic system due to Russell and Whitehead so let us see how we can prove this particular kind of theorem so now so depending upon what axiom that you are going to choose we can start with any one of these axioms and all so if you want to show that this is true this is a theorem then one needs to start with one of these things because these are the only things which are given to us it seems that the third axiom to take and take into consideration then somehow we will get into this particular kind of form not B implies A implies B so this is axiom number three you need to provide justification on the right hand side or maybe here the left hand side alone so now one instance of this particular kind of axiom is this one so now what you have done here is is that for a you substituted not be so wherever you find a you substitute it with not be a means not be so that is why not is already there that is why it becomes like this so now this is not be and a is not be and then B is as it is so this is the first step that we have so now just now so there is a law of identity which we have showed it just now that is B P is the one which we have showed just now so now in this if you put P for not be then this will become this so this is what law of identity you can say this is which we have already proved so now so what we will be using is in order to simplify this proof so not not be in place B in the case of Russell weighted axiomatic system it involves some 14 14 steps and all to show that not not be in place B is a rule of double negation so now we will pause things for a while here and then we will talk about one important theorem in the axiomatic prepositional logic so that is the deduction theorem and then now then we will make use of this deduction theorem in proving this particular kind of thing so this is what is considered to be the deduction theorem so I will come back to this particular theorem little bit later so now this deduction theorem is due to Herbrand in the year 1930s and the same time even natural deduction systems have also come into existence we do not know exactly what kind of relation you will find it between Herbrand's deduction theorem and natural deduction theorem natural proofs using natural deduction theorem that is due to Provid's and others again next so this theorem says like this of course every theorem has to find a proof suppose if you if you take ? as set of well-formed formulas in a sense that it has all the well-formed formulas which you can think of and you single out two formulas a b they are considered to be individual formulas and if it so happened that b is deduced from ? and a then a ? b can be deduced from ? so that is like this so this is what we discuss about it so now you started with a particular kind of set of well-formed formulas now then from that you also have a and from this you deduced b so if you can deduce b from this thing that means b has come after some kind of steps some finite number of steps you got b if that is the case then you discharge these assumptions and then you talk about this thing from ? you can even derive a ? b that is what is the case this is what is called as deduction theorem that means you from a given set of formulas ? and taking an assumption a you deduce b that means you already said to have deduced a ? b from a given set of formulas ? particularly if you have this particular kind of thing a and from that you generated b this is what you write it in this way then of course ? is already there here then you say that it is ? a ? b this is the same thing which we have said already so this is what is considered to be a deduction theorem actually in mathematics every theorem has to find a proof and all at this moment we are not trying to produce proof for this particular kind of thing otherwise it has to every suppose if you say that it is a theorem and if you do not have a proof then it is not considered to be a theorem so every theorem has to find a proof but due to the limitations of time and all we are not going into the details of proof of this particular kind of theorem but we make use of the idea behind this particular kind of theorem so there are two important corollaries for this particular kind of theorem so they are like this suppose if a ? b and b ? c are there already there and one of the outcome of this one is is that you can deduce a ? c so you have you have deduced let us assume that these are the two hypothesis and all so now let us try to prove this thing you have a set of formulas ? and then you have a ? b and you have b ? c and from that you will get a ? c so how do we prove this thing you take a ? b as a hypothesis and b ? c as another hypothesis now you assume the antecedent of this conclusion let us say this is the conclusion aim ? c so now you take the antecedent of your conclusion which appears in the form of a condition so these are the steps that we have so now one and three modus ponens one and three modus ponens you will get b so now in the fifth step to and for modus ponens that is this principle we have used from a if a ? b then you can deduce b so now these two modus ponens you will get C so now in the natural direction proof and all or you can use direction theorem now since from from a you got C so that means it is like this so from C C is obtained from a with some kind of steps in all one or two steps are there involved in this thing that means you can write here in the sixth step you can write like this now in the seventh step you can simply write like this this left hand side goes to the right hand side and you will get C so a ? c is the one which we are trying to deduce so one of the important corollaries of this particular kind of theorem is that if a ? b is reduced and a ? b and b ? c is the hypothesis then from that you can deduce a ? c so this we can make use of it and the other important color corollary is missing from a ? b ? c and you have b then you can deduce a ? c so this is in this kind of thing so this is one of the important corollaries of deduction theorem so what is that we will write it down here a ? b ? c a ? b and then a ? c so this is another important corollary and then I will go into the proof of so these are the two corollaries corollary one and then this is so now you have a ? b ? c a ? c sorry a ? b from that you can deduce a ? c so now let us see how we can do it so now first thing which you will find it in the slide is that a ? b ? c is what is given to us so now a ? b is already given so there are the two things which you find it in the hypothesis and these are the given kinds of things so now using axiom number two that is a ? b ? c is a ? b ? c so that is the axiom that we will make use of it and then you apply modus ponens on one and three because you have the same thing a ? b ? c then what you get is a ? b ? a ? c so now we already have a ? b so that means you will get a ? c if you want to show it clearly and all so now I think you do not have space here so we will not go into the details of that one this proof is already there here so now this is corollary to now we make use of these theorems in proving this ? b so now this is in this particular kind of format so for example if you take into consideration this as a and this as b the whole thing and this as c it is like this thing a ? b ? c a ? b ? c so now the second statement actually that is the important corollary of this one is like this suppose if you have a formula like this a ? b ? c and then b then you will get a ? a ? c so this is also one of the important corollaries of direction theory so that is the theorem which we have to do not just simply b from that you will get a ? c so now in this sense now you take this a as this one the whole thing and b as this one c as b so now we have a formula like this a ? b ? c the first formula and then b is same as this one this particular kind of portion and from that you should be able to get a ? c so that is so what we should get here if you apply this particular kind of thing so this a ? c is the one which you need to get what is a here this is not be implies not not be implies c is b so that is what you get by using corollary to actually this needs to be modified in this particular kind of sense a ? b ? c and b from this you will get a ? c one can prove it by one can show it by using set of things which we already know one of these axioms you can take into consideration and maybe modus ponens etc you apply on this one you will get this a ? c for this you start with a ? b ? c as a first step and then you assume this second thing and then third thing is you assume the antecedent of your conditional that is a so now as a fourth step one and three one and three modus ponens you will get b ? c so now fifth step b and two and four again modus ponens that means this one b and b ? c you will get C so now you have deduced c from a so that means you apply direction theorem again and this will become a ? c means this is a ? c this is the way we can prove this particular kind of theorem corollary will come as an outcome in this way so you not have to apply any axiom here you just use modus ponens rule and ultimately you got this particular kind of thing the same rule is employed here and this is now what we get so now till now it is not in this particular kind of format now somehow we need to use axiom number one a ? b ? e if we can substitute a for not not a for not not b and b remains as it is then for a it is not not b implies b is as it is then a is not not so now observe this particular kind of thing not not b implies b implies this one and the same thing not one second b as substitute b as not so then it will remain the same thing so now not not b implies this one some x and this x implies this thing now use corollary one so that is if a ? b and b ? c a ? c is the case so you have to read it in this way from 6 to 7 you need to go not not b implies this one and the same thing implies this that means this will become not not b implies b this is what we are trying to show so now using this deduction theorem and it is important corollaries we might simplify our proofs so again what is this deduction theorem again it preserves the truth so every step of your proof is a kind of truth preserving kind of thing which we are employing here so that is why the final final step of your proof is also considered to be a theorem so final step of your proof is usually considered to be a theorem so that is why this is proved in this particular sense so there are some other important and interesting proofs one can do it this is this is only for our practice so the more and more we practice the more and more efficient we will become in deriving these theorems so now let us say we are trying to prove instead of not not b ? b we are trying to prove b implies not not b so this is the other way round bc not not b ? b is a double negation but we are trying to show this so how do we go about this thing so instead for proving this particular kind of thing you need to choose some of these axioms so they are like this first you start with axiom number three what is this axiom number three not b ? not a not b ? a ? b of course you can do the same thing by using its corresponding axiom and all which is there on that is not b ? not a ? b one can use this one also but we are making use of revised version of Hilbert Ackerman system so now you start with this particular kind of thing now you substitute not not b for wherever you b occurs you substitute with this thing and a wherever you find a you substitute with b then this will become what is b now not not b so not of what is a here is b so b means not not b this a means b and now b means not not b what is this instance of instance of axiom 3 this is what is considered to be an instance of axiom number three so what is that we are trying to derive we are trying to derive b ? not not b so you might ask we might ask ourselves that so I need to follow all these steps now I can jump to I take one axiom and then jump to this particular kind of thing usually you do not get it like that so it is a path which leads to this particular kind of truth one truth is leading to another kind of truth so this is not it over somehow we trim this axiom in such a way that the last suppose if you take this as a whole well form formula the last part of this conditional is somehow turning out to be this one it is coming closer to this one so this is the second step so now just know we showed law of double negation this is what we have already proved this is what is double negation so now in this one you substitute for b not wherever b is there you substitute with not b then it will become and b is this one so now for this is what instance of double negation so now fifth one two and four modus ponus because this is same as this one these two modus ponus you will get not not be implies b that implies whatever is it not so now till now we did not get this thing and all somehow we need to use some other kind of axiom and we need to convert it into appropriate form so now we have this axiom a-b-a so now here in this one suppose if you can somehow convert this b-a as the same thing then you can say this particular kind of thing so in this one what you do is you substitute wherever a is there you substitute with b and wherever b is there you substitute it with not not b so now this will become instead of a you have b here and b means not not not b implies this b is as it is b as one second is that we are trying to do some of this needs to be converted into this particular kind of one second so you use as it is only b is same as not not b so this will become this one and a as one second this is a implies b implies a not not a as not not b so now this axiom will become this one second so what you have done here is this thing this will become not not b implies not b implies a means not not so what you have done here is that for a you substituted it not not b and for b you substitute with not b so now this is what it becomes so now what we have here is this thing I am sorry here very sorry for this so it is a implies b implies a this is axiom number one and somehow this should be converted into this particular kind of format so now for a if you can take as b and then b as not not b and then a will become this simple kind of translation and all so I am just so what you have done here for a you substituted it with b wherever a is there you substitute with b and for b you substituted not not this is what happens so now this is the seventh step so now observe this two things this one and this one so now this is b implies some x and this x implies not not b so that means using corollary one of course this follows here we can say that it is b implies not not b why it is the case because b implies not not not b implies b but the same thing not not not b implies not not b so then this goes to this particular kind of thing so that is what we are trying to prove so in this way we can prove b implies not not b how did we do this thing is started with an axiom and then again we use one important corollary of deduction theorem then our proof has become simplified here so let us consider one or two more proofs and we will end this lecture is only for our practice we are trying to talk about more number of proofs mean proofs of theorems so now this is what is famous kind of instance of material implication so now this is what you are trying to show so from not a implies the follows so now how do you prove this thing first you list out this is considered to be one hypothesis that you write it like this and then you take a also as hypothesis then we have so we have an axiom that is a implies b implies so this is what is called axiom number 1 and if you transform this thing into certain way if you substitute not b for a this is also considered to be an instance of axiom number 1 is the fifth step so now you apply modus ponens on these things you will get these particular kinds of things so now another instance of this particular kind of axiom is this implies suppose if you substitute not a for a and not b for b in axiom number 1 so you will get not b implies not a this is also in instance of axiom number 1 so ultimately we need to show that b should come as an outcome of this particular kind of thing so this is what is also an instance of axiom number 1 so now 2 and 6 2 and 6 modus ponens what is 2 here not a and not a in place this one this modus ponens you will get this thing 2 and 6 modus ponens you will get this so now and 3 and what else is the thing here 3 and 5 modus ponens you will get so we listed out the hypothesis in this conditional not a and we also assume that a is the case so now we are trying to show that b follows this particular kind of thing if that is the case then we can show that not a in place a in place b is the case so now till now the proof is not it over so we have generated not b in place a and we have not b in place a not a so now the axiom number 3 is like this not b in place not a is same as not b implies a implies b what is this this is axiom number 3 so now observe this 7 and 9 is 7 and 9 modus ponens again you will get this particular kind of portion not b in place a implies b under the 10th step so now observe 8 and 10 so 8 and 10 again modus ponens 8 and 10 is here not b in place a and not b in place a less from these two you will get b because the same as this one this gets detached and what you get is this one b so this is not the end of the proof and all so what we essentially show is this thing so I am writing it here so now what is that we got from a and not a what you got here b so now this is what we have showed so now we need to apply deduction theorem twice so that these two things will come and at the right hand side so now first time when you apply deduction theorem this goes to the right hand side so now there is an order which you need to follow suppose if you have two formulas a not a and b first time when you apply modus ponens this goes to the right hand side this will become not modus ponens deduction theorem it goes to the other hand and then it will become a implies b so now next time when you apply the same deduction theorem it will become not a implies this goes to the right hand side it will become a implies b so like this one can use deduction theorem two or three times and all I can move all the left hand side things to the right hand side so this is the way to prove famous paradox of material implication this is one of the instances of paradox of material implication so this is the way to show it so now just we will get ourselves familiarized with this deduction theorem for example if you have set of formulas like a not a b and that you get C this is what you obtain from let us say this C is obtained from these three things so now you keep on applying deduction theorem for the first time when you apply it this goes to this particular kind of thing then it will become a implies C so now the next time when you apply this particular kind of thing of course gamma is already there set of well-formed formulas taken together with these things leads to C next time when you apply deduction theorem this goes to the other side then it will become not a implies C so now if you want to eliminate this also then you need to apply this is second time you applied first time and deduction theorem applied third time leads to this so this goes to the right hand side this will become B implies not a implies C so whole thing is in brackets so like this one can use deduction theorem n number of times then ultimately this one can show it as a tautology whatever is there in the right hand side suppose you write it like this there is nothing at the left hand side that means whatever follows after this one is considered to be a theorem so that is what we can one can show so now let us consider the last kind of a theorem is called as law of contra position by using same kind of Hilbert Ackerman axiomatic system so with this I will end this lecture so what is this law of contra position that is not B implies not A implies A implies B so in each and everything which you are trying to prove what essentially one requires is that what kind of axiom one needs to take into consideration so that that goes as far as possible closer to this particular kind of thing so again we make use of this thing the last axiom a not B implies A implies B this is axiom number 3 so now we are proving it here with the help of this thing now take into consideration some of the hypothesis so that is not B implies not A as your hypothesis so this is the suppose if you assume that this is the whole conditional and all so now the first part is considered the antecedent and this is the consequence so now in this antecedent part is assumed so that is not being plus A and you also assume this particular kind of thing that is not A is also considered to be hypothesis or assumption etc now from this you need to prove B so not B implies not A of this particular kind of thing which we assume so what essentially we have done here is that first we started with this hypothesis that is the antecedent part of your conditional that is considered to be assumption or hypothesis now we have used this particular kind of axiom so now this needs to be stated below that but it does not matter let us say this is the second step in the third step not being plus not A and not being plus not a same thing this by modus ponus 2 and 1 modus ponus you will get not B A implies B this is still not in this particular kind of format A ? B we need to do a little bit of this thing now we make use of axiom number 1 that is A ? B ? A this is axiom number 1 so now one instance of this particular kind of axiom is like this A implies for B you substitute with not B and this will become like this so now this is instance of axiom number 1 so now we have A ? not B ? A and not B ? A ? B that means in the sixth step you will get A ? B so how do we get this one this X ? Y and Y ? Z that means X ? Z that is A ? B so now what we have shown here is that from assumption not B ? not A you prove A ? B so now you apply deduction theorem here then this goes to the right hand side and this will become not B ? not A ? A ? B so this is what is considered to be law of non law of contra position so in this lecture what we did is simply like this that we presented Hilbert Ackermann axiomatic system both in the unrevised format and revised form you are taken into consideration the revised form H ? which consists of the third axiom this one otherwise it is it was like this but not B ? not A is nothing but A ? B so that also you can take it as one of the important axiomatic system so if we presented the axiomatic system which involves only implication and negation signs and then we use transformation rules and modus ponens then we derived some of the theorems and we also made use of one of the important theorems of axiomatic prepositional logic so that is the deduction theorem direction theorem in a nutshell tells us that if we have set of formulas ? and your formula A and from that if you have deduced B then you are also said to have deduced A ? B by using the same set of formulas ? and there are two important color is that we have discussed in greater detail that is we also showed we also proved these things suppose if A ? B and B ? C as hypothesis and then from this A ? C will come as an outcome so that is a kind of rule of syllogism and so another important property is that if A ? B ? C is the case and B is the case then one can even deduce A ? we made use of this color is and deduction theorem and then we have simplified the proofs that are there in the given axiomatic system so far we have studied principia mathematical Russell axiomatic system due to Russell Whitehead and another axiomatic system due to Hilbert and Ackerman in the next class what we are going to see is that are these systems complete that means in a sense that all the probable things that are there in this axiomatic systems are true a valid or all the valid formulas are considered to be true whether or not the system is complete etc we will establish these things by using by making use of some kind of meta theoretic theorems such as theory of consistency theory of soundness etc. In the next class we deal with whether or not principia mathematical is complete etc all these important questions will be dealing with in the next class.