# Mod-01 Lec-31 Hilbert and Ackermann System

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
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
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
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
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
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.