Welcome back in the last class under section

axiomatic propositional logic we presented Russell whiteheads axiomatic system that is

what we find it in the book principia mathematician, in one of the sections on directions 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? Our main goal was this that all the

valid formulas in your formal axiomatic system should find a proof.

So this 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 to set of mathematicians

the Hilbert and Ackerman Hilbert is famous for his different challenges in all hell but

they are called as Hilbert problems, so we will not be talking about all the problems

in 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 exemitizing

geometry, so he lived from 1962 to 1943 his dream was indeed to create a forfeit axiomatic

system for mathematics or at least now if you restrict our self to at least geometry

in a thematic as far as geometry in a thematic 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 probable

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 finite methods which could not lead to any contradiction start with tautologies

you never end up with a contradiction. Then this part can be used as secure foundations

for the entire mathematics, so you showed that by using some kind of finery methods

we 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 theorem and all.

So it is only till Gödel, Gödel has come up with an interesting theorem which is called

as in Campinas serum account into which for any formal axiomatic system as is the case

of a principia mathematician are the hill Whitaker axiomatic system. So he has come

up with a radical kind of view that is that, there is there are always some statements

about natural numbers that is in the automatic 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 in incompleteness

what is incompleteness anything which is provable 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 Gödel has come up with an interesting

kind of theorem with which one can no one can show that no consistent system can be

used to prove its own consistency. If you think that you know principle mathematician

and Gilbert Ackerman system are considered to be consistent anyway Gilbert’s dream has

been shattered by this one of the important theorems in logic that is Gödel theorem,

which we will talk about it under the limitations of the first order logic. So as far as propositional

logic are concerned they are all decidable and complete and consistent and sound, but

whereas in the case of first-order logic that semi decidable and is too incomplete as a

triangle and all. So in the context of first-order logic when

I talk about predicate logic these things will become a prominent. So our goal is to

present in this lecture is to present Hilbert Ackerman axiomatic system and then we will

be proving some important theorems such as P ?P P are ?ˆ? be accepted and then not only

that thing we will be making use of one important theorem propositional logic axiomatic propositional

logic. So that is the de deduction theorem, if you

can use direction theorem with a set of axioms that you already have then our proofs with

our proofs will become simpler. So now to start with the Hilbert Ackerman

axiomatic system it is presented in various ways in various textbooks in particular in

most of the test books these are the three exits that are provided in the textbooks,

the like Mendelssohn etc. So these are some of the axioms you have to note that Hilbert

Ackerman makes use of only two primitive logical symbols that is implication and negation,

so there is only two symbols which you commonly find it in you find it in the Hilbert Ackerman

axioms. So to start with any axiomatic system consists

of a set of axioms and transformation rules and of course the rule of detachment, so instead

of five exams as is the case of veteran Russell invited we have only three exams here and

all these exams are in the are expressed in terms of implication and negation. The axiom

is A ?B impress a and the second one is B ? C ?A be blessing ?C and now this third axiom

for stated like this in the beginning ? being pressed ? A ?B.

But in some textbooks like Mendelssohn introduction to mathematical logic you will find this particular

kind of thin ? A + B ? A ?? B is but 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 exams one and two and the fourth one which is there at the top of the

slide, then you can formulate in a different kind of axiomatic system that is 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 naught B + naught A ? B will get it as an outcome of this revised

kind of axiom that is naught A naught B ? naught C ?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 in the tradition of

Hilbert Ackerman axiomatic system. So now using these axioms or essentially we

are trying to do is we are trying to prove some important theorems, so now so far we

have seen some important theorem such as P ?P in Russell by trade axiomatic system. So

now we try to show with the help of Hilbert Ackerman axiomatic system, you will be proving

some of the important theorems such as A ? A. So 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 Ackerman and then this is the

axiom number one in place team Ha2 A ?B ?C is a ? C, so third one is this ? B ? not if

this is what we are going to take into consideration the revised axiom this means ? A ? B. this

brackets needs to be closed purple in you want I 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 is 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 reduction theorem,

so which you 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 ? this detachment route.

So now what exactly we are trying to do is we are trimming these accents 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 proposition. So now

for this you start with the paper Tuckerman axiomatic system to that is A ?B C same thing

which we are writing it again is like this is like A? C.

So now in this what you will do is for wherever you find be you substitute it as A? C and

wherever you find C to substitute with A and of course even B also C. So that means you

are replacing B with A and you are replacing sorry you are replacing B ? C with A and this

is not required, so these are the two operations that you are trying to do.

So the rationale behind this thing is that if you substitute anything into this axiom

uniformly that will retain is tautology 0, 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 easy

A as it is because we are proving A?A that means you have to eliminate this B’s and C’s

somehow, so that you will find only a formula. So that is the reason why we use this uniform

substitution rule. So now what is B here B means ? A and then

C is also considered to be A now second A?A A?B is am passing, so that is the first one

and then second one is A ?B we ensured that the last step of your condition lost the consequent

of your conditioner at the occupying the last position is somewhat closer to what we are

trying to prove, so let me say impress you. 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 decide. So now this is step number two 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 is axiom number one Ha1 if we can understand one proof then we can

solve we can prove many other theorems in. 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 Ackerman and dramatic

system one that means you substituted 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 2 and 3 you have to write justification here more respondents

that means we use this particular kind of rule then this gets detached and then whatever

remains is this one in the same place is so that is A ?A ? A this part goes in on it gets

detached and whatever remains is this portion whatever is there afterwards A?A.

So this is how did we get this one by applying modus ponens on 2 and 4, so now in the sixth

step still it is not in this particular kind of form somehow you need to detach this, how

do we detach this particular kind of thing, somehow we need to again fall back on or axioms,

if we 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?C, so now A ?A

? B what is this is an instance of axiom number one, because instead of B we have to put A

here uniformly you substitute A for B, so this is what you get. So now under the eighth

step so these two are same A ? A impressive and impressing impressive this is like X and

X ?Y so you will get what. 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 the 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 you 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 serums might come in four steps sometimes seven steps sometimes, if you are axiom the

choice of your accents are wrong then you will you will be playing with it and ultimately

it might it might take some sixteen steps sometimes proofs might take even days also.

So the effective proof is considered to be that particular kind of proof which ends in

final steps infinite intervals of time that means if you prove never ends in it goes on

and on and all 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 you give the feedback of all these axioms in all and then

you put this data A ?A say whether it is a theorem or not that software will tell us

whether this is a theorem or no or there are some soft waves which provide even proofs

also that is not what we are going into the tails of it.

So as a first step we are trying to show how we can generate a in A ? A by producing some

kind of rigorous proof, so with this proof 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 tooth preserving rule and

transformation rules also preserves the troop. So everywhere we are every way the you considered

as hypothesis 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 set 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 is

A, so this also can be proved in all if this is rude then ? A can also come as a B. So

now what we will be doing is, so we will be proving some other kinds of theorems such

as ?b. So let us get our self familiar 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 consists of only few rules and using this only 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 in all so if you use anything outside the things and on like in the case

of Euclidean examining system it 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 increase it assumptions which are part and parcel of

your proof and there are certain things which are not part of the proof also tell they also

took part in the proofing. So in that sense 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 that non rigorous kind proof.

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 really how we have already proved it in the in the axiomatic system due to Russell,

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 in

all, so if you want to show that this is true is this is a freedom 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 ?B? A ? C. So this is axiom number three you need

to provide justification on the right hand side or maybe here the left hand side, so

now one instance of this particular kind of axiom is this one. So now what you have done

here is that for A you substituted ?b, so wherever you find a you substitute it with

? A means ?b, so that is why ? is already this that Is why it becomes like this.

So now this is ? B and is ? B and then be is as it is so this is the first step that

we have one and two, 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 a PE for not be then this will become this

so this is what law I entity you can say is which we have already proved. So now so what

we will be using is in order to simplify this proof so ?B?B in the case of for wrestle weighted

axiomatic system it involves some 14 steps and all to show that ?B?B be 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 propositional logic,

so that is the deduction theorem and then now then we will make use of this direction

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 a little bit later, so now this deduction

theorem is due to Herbrand in the year 1930 and the same time even natural direction systems

are also coming to existence, we do not know exactly what kind of relation you will find

it between Herbrand deduction theorem and natural deduction freedom natural proofs using

natural deduction theorem, that is due to profits and others.

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 form 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 in A ? we

can be reduced 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 did used B, so if you can reduce be 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, so that is what is the case this is what is called as

deduction theorem. That means say from a given set of formulas

? and taking an assumption a you deduce B that means you already I already said to have

reduced in place be from a given set of formless ?, 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 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 on at this moment

we are not trying to produce proof for this particular kind of thing, otherwise it has

to 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

in 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 coral reefs for this particular kind of theorem, so they are

like this suppose if A ?B & B ?C are there already and one of the outcome of this one

is that you can deduce A ?C, so you are reduced let us assume that these are the two hypotheses

are not so now let us try to prove these things you have a set of formulas ? and then you

have A ? V and you have B ? C and from that you will get A ?C, so how do we prove this

thing you take A? Bas hypothesis and B ?C as another hypothesis.

Now you assume the antecedent of this conclusion let us say this is the conclusion A?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 modes

one is you will get be surround the fifth step 2 & 4 modus ponens that is this principle

we have used from a if A ?B then you can reduce. So now these two modus ponens will get C,

so now in the natural direction proof and on or you can use direction theorem now, since

from A you got C, so that means AC like this so from C is obtained from a with some kind

of steps in on one or two steps are there involved in this service you can right here

in the sixth step you can write like this. Now in the seventh step you can simply write

like 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 & B emphasis

hypothesis then from that you can deduce A ? C. So this we can make use of it and the

other important corollary is this thing 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 corollary

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 this

thing so these are the two corollary 1 and then is corollary 2, so now you have A ?B

?C and K imply see sorry A ?B from that you can reduce ?C, so now let us see how we can

do it. So now first thing you will find it on 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 same ? B ?A ?C, so that is the

axiom that we will make use of it and then you apply more respondents on one and three

because you have the same thing A ? B ? C. Then what you get is A ? B let us say impressive,

so now we already have A? B, so that means you will get A ?C, if you want to show it

clearly in all so now this I think you do not have space here so will not go into the

details of that once proof is already there here, so now this is corollary to now we make

use of for these theorems improving this ??B?B. So now this is in this particular kind of

format, so for example if you take into consideration this as a and this has be 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 be then you will get A?C.

So this is also one of the important corollary soft direction B so there is the theorem which

we have protein off will just simply be from that you will get A ?C. So now in this sense

now you take this a as this one the whole thing and be 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 ?B? ??B ?C is be 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 except I we apply on

this one you will get A?C, for this you start with the A?C as the first step and then you

assume this second thing and then the third thing is doing the antecedent of your conditioner

that is A, so now as a fourth step one and three more respondents you will get B ? C.

So now fifth step B & 2 & 4 again more respondents that means this one B & B in blessing 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 in ? C means this is ? C this is the way we can prove this

particular kind of theorem the corollary will come as an outcome in this way. So you not

have to apply any axiom here you just use modus ponens rule then ultimately you got

this particular kind 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 some

of you need to use axiom number 1 A ?B ?A if we can substitute A for ?? if ? our B & B

remains as it is then for a it is ??B?B is as it is then is ?,

so now observe this particular kind of thing ??B in place being bless this one and the

same thing not one second and P as substitute be as not, so then it will remain the same

thing so now ??B? this one some X and this X ? this thing.

Now use corollary 1, so that is if A ? B & B ?C name place C is the case, so you have to

read it in this way from six to seven you need to go ??B ? this one and the same thing

?this that means this will become ??B ?this is what we are trying to show. So now using

this deduction theorem and it is important corollary is we might simplify our proofs,

so again what is this direction 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 step of your proof is also considered to be 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 will become in deriving these theorems.

So now let us say we are trying to prove instead of ??B in place means we are trying to prove

be B? ??B, since the other way round we see ??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 thing what is this exit number three, ?B in ? A, ??B of course you can do

the same thing by using its corresponding examine all which is they ? A ??B ? A ??B

one can use this one also but we are making use of revised version of a Hilbert Ackerman

system, so now you start with the this particular kind of thing now you substitute ??B for wherever

you be occurs you substitute with this thing and A wherever you find a you substitute with

B. Then this will become what is B now ??B so

not of what is A here is big so B means ??B means and now b means ??B what is this instance

of instance of axiom three this is what is considered to be an instance of exit number

three, so what is that we are trying to derive B? ??B. So you might ask we might ask ourselves

that so why you need to follow all these steps now I can jump to I take one axiom in and

jump to this particular kind of thing usually you do not get it like that so it says it

is a path which leads to this particular kind of truth one true three is leading to another

kind of true. So this is not it over somehow we trim this

axiom in such a way that at the last suppose if you take this as a whole well formed formula

the last part of this conditioner is somehow turning out to be this one it is coming closer

to this one, so this is the second step. So now just now we showed love for double negation

this is what we have already proved this is what double negation is. So now in this one

you substitute for be not where ever be is there you substitute with ??B then it will

become only and B this one. So now for this is what instance of double

negation, so now fifth one 2 and 4 more respondents because this is same as this one is two more

respondents you will get ??B ? B that ?whatever is a ??B, so now till now we did not get this

thing you know some of you need to use some other kind of axiom and we need to convert

it into appropriate form. So now we have this axiom A?A so now here in this one suppose

if you can somehow you convert this being classy 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 then we substitute with B and where ever B is there you substitute it with

??B, so now this will become instead of a how B here and B means ??B B ? A B as one

signal is that we are trying to do some of this needs to be converted into this particular

kind of thing once. So you use as it is only be is same as ??B,

so this will become this one and ??B, A ??B, so now this axiom will become, so what you

are done here is the same this will become ??B ? ?B ?A means ??B. So what you are done

here is that for A you substituted it ??B and for B you substitute with ?b, so now this

is what it becomes. So now what we have here is this thing I am sorry here just very sorry

for this so it is A ?B ? A this is axiom in 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 ??B and then A will become the same there is a simple kind of translation not so I am

just so what you have done here for A you are substituted it with B where every is there

you are substituted with A and B you have substituted ??B 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 ? some X and this X ? ??B. So that means using corollary 1 because this

follows here you can say that it is be in place ??B why it is the case because B ? ??B

?B but the same thing ??B goes to this particular kind of thing so that is what we are trying

to prove. So in this way we can prove B ? ??B, how did we do this thing you started with

an axiom and then again we use one important corollary of direction 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 ? in A?B 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?B ? A. So this is what calling axiom number

one is and if you transform this into certain way if you substitute ? B for A this is also

considered to B an instance of axiom number one 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 for this particular kind of axiom is this so A? suppose if you

substitute ? A and ??B B in axiom number one, so you will get ?B ??A is also in instance

of excellent number one 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, so now 2 and 6, 2 and 6 modus ponens what ? and ??this one this modus ponens you

will get the same 2 and 6 modus ponens you will get this. So now 2 and 4 and 3 and what

else is the thing here 3 and 5 3 and 5 modus ponens you will ??A. So we listed out the

hypothesis in this conditional ?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 ?B? B is the case. So now till now the proof is not it over,

so we have generated not being pressing and we have ? B proceed not yet, so now the axiom

number three is like this ?A ??B ?A?B what is 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 potion

?B?A?B another 10 step, so now observe 8 and 10.

So 8 and 10 again more response a 10-10 is here not being present and ??B ? 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 at 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 eight and

?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

at the right hand side, so now first time when you apply detection 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 and ?B first time when you apply modus ponens this goes to the

right hand side this will become not modus ponens, direction theorem it goes to the other

hand and then it will become A?B. So now next time when you apply the same direction

theorem it will become ?A in place this goes to the right hands and it will become A ?B,

so like this one can use direction theorem two or three times and all I can move all

the left hand side things to the right hand side, So this is a way to through famous paradox

of material implication this is one of the instances of paradox of material implication.

So this is the way to show so now just we will get our self familiarize with this deduction

theorem example if you have set of formulas. Like A? and ?B and then you get C so this

is what you obtain from let us say this C is obtained from these three things, so now

you keep on applying direction theorem for the first time when you apply it this goes

to this particular kind of thing then it will become a nice C so now the next time when

you apply this particular kind of thing of course ? is already the set of well form formulas

taken together with these things leads to C next time when you apply direction theorem

this goes to the other side then it will become ?A ?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 in place what is that B ??A?A?C the whole

thing is in brackets. So like this one can use direction theorem and number of times

then ultimately this one can show it as a tautology whatever is in the right hand side

suppose if you write it like this there is nothing at the left hand side.

That means whatever follows after this way he is considered to be a 0, so that is what

we one can can show. So now let us consider the last kind of theorem is called as law

of counter position using same kind of a vertical axiom system, so with this in this lecture,

so what is this law of contribution it is ?B ?? in size a nice, so in each and everything

which you are trying to prove what essentially one requires is that what kind of axiom or

needs to take into consideration. So that goes as far as possible closer to

this particular kind of thing, so again we make use of this link the last axiom a ?B

?A ?B this is exam number three. So now we are proving it is with help of facing now

taking into consideration some of the hypothesis so that is ?B ? ?A as your hypothesis. So

this is the if suppose if you assume that this is a whole conditional and on, so now

the first part is considered the antecedent and this is the consequent.

So now in this antecedent part is assumed so that is not being placed and you also assume

this particular kind of thing that is ?B nice it is also considered to be hypotheses our

assumption etc, now from this you need to prove be so ?B ??A of this particular kind

of thing which we as you. So what essentially we have done here is that first we started

with this hypothesis that is the antecedent part of your conditioner that is considered

to be assumption our hypothesis. Now we have used this particular kind of axiom,

so now this needs to B stated below that but it does not matter let us say this is the

second step in the third step ?B placed ? and ?B ?A same thing this by modus ponens A?B,

this is still not in this particular kind of format A ? B we need to do a little bit

of for this thing. Now we make use of axiom number one that is stay in place B ? A this

is axiom A number. One so now one instance of this particular

kind of axiom is like this A ? B you substitute with and this will become like this, so now

this is instance of I will say number one so now we have in place ?B ? A and ?B ? A

?B that means in the sixth step you will get A ?B how do we get this one this X ?y and

y ? Z that means X in place z that is A ?B, so now what we have shown here is that from

assumption ? B equalized ?A A ?B. So now you apply deduction theorem here then

this goes to the right hand side and this will become ?B in place it ?A ?B, so this

is what is instead to be law of contract position, so in this lecture what we did is simply like

this that we presented Hilbert Ackerman axiomatic system both in the unrevised format and revised

form we are take into consideration the revised form H prime, which consists of the third

axiom this one. Otherwise it is it was like this but not being

place ?A is nothing but A ?B so that also you can take it as one of the important axiomatic

system, so 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 propositional

logic. So that is the deduction theorem direction

theorem in a tells us that if you have set up formulas gamma and your formula A and from

that if I reduce B then you are also said to have deduced in A?B by using the same set

of formulas ? and there are two important is that we have discussed in greater detail

that is, we also showed we also prove these things suppose if A ?B & B ?C has a hypothesis

and then from this A ? C will come as an outcome. So that is a kind of rule so another important

properties is that if A ?B ?C is the case and B is the case then one can even reduce

in place, we made use of this corollary send deduction theorem and then we have simplified

the proofs that are there in the rain the given axiom system exists. So far we have

studied principia mathematica Russell axiomatic system due to wrestle and another T system

due to Gilbert in Ackerman. In the next class what we are going to see

is this that are these systems complete that means in the sense that all the probable things

that are there in this exhibit systems are true a valid are all the valid formulas are

considered to be true whether or not the system is completed etc. We will establish these

things by you by making use of some kind of met theoretic theorems such as theory of consistency

they of soundness expand. On the next class we deal with whether or not principia mathematica

is complete etcetera all this important questions will be dealing with in the next class you.