Welcome back in continuation to the last lecture
where we discussed about the semantics de
blocks method we discussed some examples we
discussed some examples and we discussed about
when they are considered to be when they are
going to be valued etcetera so we will talk
about some more examples in this class so
that we will get ourselves familiarized with
this particular kind of technique so this
technique occupies the central position for
this course.
So that is why we are spending a little bit
of a little bit more time on this particular
kind of method so as I said in the last lecture
semantic tableaux method is all about finding
some kind of counter example suppose if you
are trying to check the validity of a given
well-formed formula of a predicate of predicate
logic what we are trying to do is that first
you negate the formula and then you construct
a tree based on the tree rules that we have
discussed in the last class.
And then if the negation of the formula leads
to the branch closure then we said that negation
of that formula is unsatisfiable and whenever
not X is unsatisfiable X is considered to
be valid so that is one thing which we have
been doing and then the second thing is that
if you want to talk about consistency of a
set of statements in the predicate logic then
what we need to do is you construct a tree
diagram for these two sentences and then when
the branch does not close or not that means
it satisfies the formula the given formulas
and hence these two formulas are said to be
consistent.
For example if you want to talk about consideration
px and qx and then the another thing for al
xs px ? ? qx let us consider that let us assume
that these are the two statements that are
given to you so now we would like to see whether
these two are consistent to each other or
not using semantic tableaux method so the
first thing we need to do in the semantic
tableaux method is this that always handle
the formula which consists of existential
quantified.
So now first you eliminate this quantifier
using this particular kind of rules suppose
if you have a formula like this in your tree
then if you remove this existential quantifier
then you are replacing it with some kind of
parameter A and then each time when you remove
this existential operator exist is still quantified
you have to use a new parameter which is new.
So now if you remove this particular kind
of things then this will become P a ^ Qa so
this is one of the instances of this particular
kind of formula so now we are checking for
the consistency existence here for these two
formulas now fourth one so now this px ? ? qx
holds for all X in particular so that is why
it holds for even this particular kind of
thing else one instance of this one could
be even this s ?q a.
So now we use the same rule X ? Y is not exempt
why so you apply this particular kind of rule
for this one and this will becomes ? Pa and
? qa so now P and Q a can be written in this
sense I am just writing it here itself qa
whenever you have two formulas like this P
and Q the tree diagram for this one is simply
this P and Q it looks like a trunk so P and
Q followed by that you were to write like
this in the tree diagram.
So now in this one you have Pa here and ?Pa
this branch closes I means these two are contradictory
to each other a literal it is negation is
found here that is why this branch closes
and there is no way in which you can go beyond
this one and you have Qa and ?Q were here
even this branch also closes so that means
you listed out these statements one after
another and you consecutively get diagram
and all the branches closes that means there
exist some X px and qX and this particular
kind of formula for all X Px ? ?qx it is said
to be inconsistent to each other.
Why because if you take both the statements
and construct a tree it leads to the branch
closure so that means unsatisfiability so
it is not satisfied satisfiable in any one
of this interpretation so because all the
branches closes so in that sense there exists
some x for all x px ? ?qx that is it to be
inconsistent to each other so you can replace
it with some kind of preposition for if P
we can replace p with some many other kind
of thing in the natural language you yourself
can see that if you state for all x px ? ?q
s and at the same time you say that there
exists some X px and QX and these two statements
are each other.
So let us consider another example and see
whether these two formulas are said to be
consistent or not so this for the sake of
understanding we I am taking these simple
examples then later I move on to check the
validity of it given predicate logical formula.
Let us say there exists some X Px ^ QX this
is the first statement and you consider another
statement such as Px ? QX and then third one
they take any other things such as there exist
some X it is not the case that there exists
some X Px so just for the sake of time being
you take these three things so now we want
to check whether these three statements are
consistent to each other or not so now you
will start constructing the tree diagram for
these things first you eliminate this existential
quantifier it will be Pa and Qa.
So now first you eliminated this one exists
initial quantifier and then one instance of
this one is going to be this one now with
now this can be written as ? there exists
some X Px is nothing but this negation goes
inside and negation of existential quantifier
will become Universal quantifier and you have
to push this negation inside and this will
become this so now we can write straight away
like this for all X ?Px.
So now here in the second one if you eliminate
this existence will quantify and then you
need to ensure that it is replaced by a parameter
which is not used earlier so A is the parameter
which is used here so we are not supposed
to use it again here next time when you remove
this existential quantifier you need to use
another parameter let us say B other than
this a is not going to be qb.
Now so now the next one is going to be this
one for all X ? Px so this is going to be
true for all X and all irrespective of whatever
you substitute whether ARB it is going to
be the case so that is very it is going to
be the case ?Pb so now Pa and Qa it can be
written in this sense qa so that is the fourth
one fourth one after another you can write
and the second one is Pb and Qb and then you
have ? Pb once again since you have P be here
and ?P be here this branch closes.
It turns out to be the case that these three
statements there exist some X Px and ? Px
and Qx there exist some X Px or Qx sorry this
is our we are not supposed to close like this
so this is going to be this one Pb ? Qb so
now so this is going to be like this is Pb
and Qb so this we expanded it and then it
will become Pb and Qb since there are connective
is this which I did not notice it so now ?Pb
and Pb closes and then this branch is open.
So this branch is open in the sense that you
have Qa here but you have Qb here so there
is no way in which you can cancel they can
close the branch that means this branch is
open so now from the open branch open branch
is the one which satisfies this particular
kind of this formula satisfies the formulas
means the values that that are going to be
there here satisfies that we that makes these
three formulas true so what are these things
when to be qb is t and ?Pb becomes T and then
both Pa T Qa T and this is going to satisfy
these three formulas that means they are going
to make these three formulas true,
So it is in that sense these three formulas
are said to be consistent to each other so
the only one thing which you need to note
that is the least out all these formulas one
by one after another and then construct a
tree diagram and if at least one branch is
open that means the formula given formulas
are said to be consistent but in the earlier
case when we constructed a tree diagram for
the given formulas all the branches closed
a means it is unsatisfiable so in this case
at least in one instance it is satisfiable
then it is considered to be these three sentences
are considered to be consistent to each other.
So this is the way to check whether it given
statements are consistent to each other are
not so now let us talk about some more examples
it is respect to validity for validity what
you will do is for example.
If you are given a formula X what you need
to do is you need to construct a tree diagram
for ? X and then if all the branches closes
that means you will end up with a contradiction
all the branches closes then ? X is considered
to be unsatisfiable and that ensures us that
X is going to be valid that is what we will
be doing in the case of checking the validity
of a given well-formed formula in the predicate
logic this is an important decision procedure
method because so in this method this also
serves as a proof in a sense that any proof
is considered to be considered to be ending
in finite steps in finite intervals of time.
So now let us consider this example for all
X for all Y for all Z are XY ? Y Z ? X Z etcetera
and the second statement is for all X for
all Y ? XY ? YX and the third statement for
all X there exists somebody X Y and the fourth
one for all X all x so now we want to show
whether 1, 2,3 in the 1st 3 statements leads
to the fourth statement or not that means
four statement is considered to be a semantic
consequence or logical consequence of 1, 2,
3 so for that what you will be doing simply
is that we list out all the three statements
one after another.
And you take the negation of the conclusion
and then you start constructing the tree and
if it turns out that all the branches closes
then the negation of the conclusion is unsatisfiable
that means the given conclusion is correct
is considered to be the correct kind of conclusion
from these three premises so now let us consider
this particular example and then we will see
so why we are doing all this thing because
you have to getting our self familiarize with
this particular kind of technique we are solving
these particular kinds of problems.
So now the statements are the given statements
in the predicate logic are like this first
one for all X for all Y for all this is the
case R X and R Y Z ? R X this is a kind of
for some transitivity property second we have
for all X for all Y we have R X Y ? RY X,
3 for all X there exists some Y R X Y, 4th
one so now this is constructed with a conclusion
for all X R X ,X so now we want to check whether
this particular kind of statement follows
from these three are not using the technique
of semantic tableaux method.
So for that what you need to do is to take
into consideration the negation of this formula
so that is ?for all X R X, X there is a different
kind of notation that is being used here sometimes
I write R X, X and some other textbooks which
is simply written as R xx so just to separate
the predicates with the individual variables
subscript and superscript and subscript you
write it in this way.
So it does not matter I know whatever way
you write R X, X means this X and X are in
some kind of order it can be written in this
sense our followed by X, X or it can be written
even in this sense so it is used interchangeably
and the other thing is that for quantifiers
in some textbooks you put parentheses like
this just to separate this one for example
you can write like this in some other textbooks
it is simply this parenthesis is ruled out
and then you can simply write X and X okay.
So this is only for our convention all, the
all these things are correct on correct kind
of correct ways of representing the same thing.
So now we are showing that 1, 2, 3 is a set
of propositions lease to 4 so that means these
three things leads to this particular kind
of thing so now for the semantic tableaux
method you start with the negation of the
conclusion so now ? for all X or X, X means
this one if you simplify this one and this
is very existence X and then you put this
negation in say it will become are X and X.
So that is what we are going to write here.
So this is they reduce some X ? R X, X so
this brackets needs to be cleared out so now
the strategy for the semantic tablets method
is that first you need to eliminate this existential
quantifier before handling the universal quantifiers,
so the best thing to remove or listening to
handle is this 5th one so that is when you
replace this when you eliminate this existential
quantifier and we have a rule if something
is true this then p ( a ) where you have to
replace this thing with p (a) where a the
parameter is new.
So now this is going to be like this ? of
R a and a the first thing that we will be
trying to do 
so now in the same way 6, 1, 7 so now coming
back to this one for all X there exists some
y all x y that means there exists some y R
X, Y that holds for all X that means it holds
for even when you substitute a for X that
also it is going to be it is going to hold
for that particular kind of thing else so
this is for all X there exists some Y or R
X Y.
So now this particular kind of thing holds
for all X that means there exist some Y are
even if you replace it X with a that is going
to hold no so now this is what you are going
to write here there exists some Y so how did
you get this one 5 existential instantiation
for all the y R so this is ay this is this
3 Universal instantiation because your remove
this Universal quantified so now eight since
this is the only thing which we have in this
one which starts with the existence quantifier
the second with this thing and then we move
on to the universal quantifiers.
So now there exists some by R ay if you remove
this particular kind of existential quantifier
you have to ensure that when you replace Y
with any other parameter that parameter should
not figure out in any one of this any other
things about this particular kind of formula
that means when you remove this y it has to
be it has to be b rather than a anything other
than a you can substitute it for this.
So this is going to be Ra b rather than ra
because a is already exhausted here so this
is 7 essential institution nine 
so now coming back to this so this is over
and this is okay now coming back to this one
for all X for all the Y R X Y ? Ryz you take
any you substitute any values X and Y that
RX Y ? R Y X holds that means if you substitute
X for a Y for B then also that is going to
hold so in that sense this is going to be
R a, b.
So you substitute X for a and Y for b and
this is going to be the case and then R y
or y to substituted b and for a substituted
a so this is R ab ? R ba so how did you get
this one to substitute X for a this is what
we have now you expand this one so whenever
you have a formula X ? Y it is like this X
? y is ? x and y so apply this on this one
it will become ? R AB and then R b a problem
is a little bit lengthy and on someone needs
to have a little bit patience to check this
validity of this particular kind of formula
so three might be a little bit big but it
still manageable it ends in finite intervals
of time.
In finite steps 
now this is what we have now observe this
particular kind of thing are a B anyhow ? R
a, b is exactly contracted to each other it
is like X and ? X now this branch closes here
itself so now we need to expand this particular
kind of branch this is this branch which is
open there is no b, a except.
So now what is unchecked is this one so we
need to note that Universal quantifiers whenever
a formula starts with the universal quantifier
you can there is no way you can check the
formula and in the case of propositional logic
for example if your P implies Q Č Q except
I and all are in place P while constricting
the semantics tableaux method first when you
are checking this particular kind of formula.
Then you write it like this and then you check
this formula like this that means you are
not you are no longer using the same formula
again but if this formula starts with universal
quantifier like this P xqx this can be used
n number of times recursively you can use
the same form because it happens for all X
so in the case of propositional logic each
time when you are expanding the tree with
these formulas you are checking this formula
so next time when you do it when you check
this particular find a formula you do like
this ČR P.
So now you take this formula and all the formulas
are checked so you put tick marks for this
particular kind of thing that is not going
to happen in this particular kind of situation
it can be used recursively not so now coming
back to this particular kind of formula now
in this formula what you do is this thing
you substitute X for a A or B and C that A
so wherever you have X is substrate with a
and wherever you have y you substitute with
B and where every A that you substitute again
with the A that substitution should B be uniform.
Then this formula will become example here
just like this for all x for all Y for all
that are xy yz implies r xz so now your substituted
like this x ? a wherever you have x cu substrate
with a why substrate with B and then wherever
you have z is substituted with A so now this
will become R now you are eliminating these
quantifiers now one instance of that one is
this particular kind of thing so now this
will become A be the first one and our yz
means instead of y we have b here B and a
implies this one are xz x means a that means
x and z are same.
So that is why are why we did like this because
we have a term R(b,a) somehow we need to eliminate
this particular kind of term so that is the
reason why we cleverly chosen these variables
to be like this so now this is what you substitute
it here now this will become so now if you
further simplify this one so this is X ? Y
so now this will become Č of Čt of R so
this is ČX ČR a B and R B a and then this
simplifies to this one R so now this is going
to be like this Č of RA B is like this not
of RAB and this is not of B and this remains
as it is so now you need to substitute the
entire thing here for this open branch.
So now we have just write it down here let
this branch remains the same so now you observe
whatever is the open branch and you list it
out on this particular kind of thing anyhow
ČRAB and then you have ČRB and then all
the way down these are the things which we
have so now observe this particular kind of
thing ČR a Č(a,a))is a branch cause actually
this should be like this since we do not have
space here so we have gone the other way around
so ČR( a,a) and then you have R (a,a) .
This closes now coming back to this one R
(a,b )Č R(a,b) this branch closes now Č(a,b)there
is something called Č R(a,b) where is this
Č R(a,b)is there here and then all the way
down here you have to write this also decipher
forgotten RBA is there and not RB is even
this also closed asleep so now all the branches
closes so what does it mean so we started
with these three formulas and then this is
considered to be the conclusion and we negated.
The conclusion and that leads to the branch
closure that means negation of the conclusion
is unsatisfiable that means X has to be valued
where it means it has to be true that means
this is the this is constructive either the
original conclusion is considered to be them
to kind of conclusion that means this follows
from these three statements so in the same
way you can check whether one in two leads
to three two and three leads to one all these
things you can check just now taking into
consideration.
The same thing that first you list out the
premises and you take the negation of the
conclusion and then see whether it leads to
the branch closure or not if it leads to the
branch closure that means the negation of
the conclusion leads to unset it stability
and means negation of x is considered to be
contradiction that means x has to be the case
x has to be 2 x has to be tautology.
So here that is a way to prove to show that
a given formula is considered to be valid
whether or not a given formula follows from
that so now let us consider some more examples
which are considered to be invalid and those
formulas which are invalid you can construct
a counter example within the domain all the
open branches indicates that it is a kind
of counter example within the domain so let
us consider some more.
Examples so that you will get used to this
particular kind of technique that is the semantic
tableaux method is one or two examples which
will be considering and then we will end this
lecture so now let us consider let us coming
back to the consistency again.
The problem of consistency let us say we have
a set like just px implies 2x and then there
exists some x px and the react is some x Č
to x so we have the in the in our set we have
for these three formulas so now we are checking
whether these three formulas are consistent
to each other or not so now you start numbering
those things 1 2 3 now you are checking the
consistency so the first thing that you do
is as usual in the symmetric tab blacksmithing
in the predicate logic is easy you are to
handle the pretty existence will quantifies.
First you can handle any one of these things
now if you eliminate this existence will quantify
here is an instance PE to existing shift this
transition now you do not have to jump to
this one now you need to handle this one so
now this is going to be not Q but you are
not supposed to use e it has to be B so this
is 3 it will change its instantiation is this
one so now 6 1 a px implies Q x holds for
all x so that is why it has to hold for pa
implies Q way it has to be true for even PB
impress QB you can you can all you can also
use that particular kind of things.
So we are used to speaking press Q a now this
is going to be pa q so PA and Č closes and
not Q a and you have not QB here and this
branch opens that means this particular kind
of interpretation satisfies this three formulas
or not that makes these three formulas two
together that means this these three statements
are said to be consistent to each of them
now if you change this problem a little bit
and then we are trying to see whether.
So now in this case these three formulas are
said to be consistent now just slightly change
this particular kind of problem and then let
us talk about the same problem unique difference.
So now let us see whether you take these two
statements into consideration.
Now whether or not this not Q X follows from
these two statements are not so now so now
we write it in the conclusion so for all xpx
implies Q X there exists some X px and then
there exists some X now not Q X whether this
follows are not from these two premises no
so how do we check whether or not this argument
is valid or not so again we use the semantics
tab Locke's method in that the first step
includes the negation of the conclusion that
is not there exist some X naught to X you
start with this particular kind of thing.
So now here we have a definition for all X
to X the same as there doesn't exist some
X naught dot X in the same way there exists
some X Q X same as not for all X not X so
the standard definitions are not so exists
a universal quantifier can be defined in terms
of existence will quantify that an existence
should quantify is defined in terms of universal
quantifier in this sense so you use this particular
kind of thing and then you put it here so
this is simply for all X cube X 3 by definition
so what we have done here taking list out
the premises.
You take the negation of the conclusion and
then we are constructing a tree and we are
going to see whether the branch closes or
not so now fifth one always handle this existential
quantifier when you remove this thing there
exists some X px it is going to be a instantiation
is this so now 6 1 you can handle any one
of these things know all these two starts
with for all X something so now one instance
of this one is going to be PE this is going
to be Čsuch P implies Q A so this is the
instance of so Universal instantiation of
this one it is going to be P a in place come
it so this is if you expand.
This one it is going to be like Č PA and
QA now you have another formula listing for
all X Q X that means it has to be true for
even a also so that's why you can write it
straight away like this or point one Q a so
this is for universal instantiation this so
now P a and not B closest but this branch
remains open that means from these two premises
a miss negation of the conclusion negation
of the conclusion does not lead to contradiction
so that means we are not able to we are able
to construct a counter example even after
denying the conclusion.
So what we in the context of in the in the
basic concepts we discussed about invalidity
an invalid argument is an argument in which
you have your premises to be true and the
conclusion is false if we can come up with
an example where premises are true in the
conclusion is false and that is considered
to be an account considered to be a counter
example for the given argument and hence that
argument is invalid so here is an instance
where you even if you deny the conclusion
you still have it still makes this satisfiable
and all that means two premises.
And false conclusion is going to be satisfiable
in this particular kind of thing especially
when QA is true P a is true then this whole
statements are going to be true that means
your two premises and a false conclusion that
will serve as a counter example so open from
the open branch you can construct a counter
example so for this particular kind of thing
you can choose a domain to be anything as
a set of people a set of reverse or anything
and then in that particular kind of thing
you need to have some kind of relation R in
particular predicate and then you whatever
is true here you list it out QA and PA are
true and based on.
That you can judge that you know you can easily
construct a counter example for a counter
example within the domain, so if we can come
up with a counter example within the domain
then obviously that argument is considered
to be invaluable so in this way we can solve
some difficult problems as well this is consider
one more example and then we will finish it
off so the in the context of for distribution
of universal quantifier we asked ourselves
whether Universal quantities are distributed
or not so that is example.
If we have this formula for all Vx Px or QX
so from this whether or not it follows that
that means whether we can derive this particular
kind of for all xpx or for all X Q X if for
all xpx in place for all xpx are refer all
explained and this is distributed over this
is distributed over distinction Universal
quantifiers are distributed over the disjunction
so again we want to see whether this particular
kind of thing holds are not so px or QX and
then here is individually we are written like
this.
So now again the using the semantic nebulous
method whether this argument is follows or
not is the one which we are trying to check
so you list out the premises like this px
or QX and then you start with the negation
of the conclusion for all x for all x so now
three you simplify this particular kind of
thing then it will become for all X px negation
of disjunction will become conjunction and
then this will be products so now this can
be written in this sense for all X px and
then if you simplify this thing it will be
like this Q X is only three simplification
you will get this one three again simplification
will get this one.
So now fourth sorry six one now you further
simplify this one not for all xpx it is not
for all X px the same as there exist some
X it puts this negation inside and it is not
the case so there exist from X ČP X and then
seventh one there exist some XČQ X so how
did we get this one for my definition and
v by definition the definition is this one
so the problem is started over yeah so now
we have there exist some X ČP X there exists
some X Č Q X and then we have this particular
kind of thing so now you always try to eliminate
this existential quantifiers first before
going to the universal quantifiers.
So now first time when you eliminate this
existential quantifier and this will become
not PA so this is six existential instantiation
and then eight seven if you apply a solution
again then this will be Č Q it should not
be a it should be B now we have this for all
xpx implies q is going to old for any anyone
any parameter you can substitute into it and
it's going to hold that means satisfies that
particular kind of formula.
So now we have Č PA and we have or not QB
and then for example if you substitute it
as PA and so one instance of this one is going
to be our two it can very well be like this
also PB r QB awesome so we take this into
consideration then this will be like this
P and Q so now in this case this branch closes
and this branch remains open but if you take
the other one into consideration instead of
PA QA you are taken into consideration P,
P and Q B.
So then what will happen is this is the thing
which happens this branch closes and this
left hand branch remains open in either cases
one of the branches remains open that means
for all xpx rqx you will not be able to derive
for all xpx are for all X Q X so now that
means that that doesn't imply this particular
kind of thing you can check whether you replace
it with there exists some X and then you can
construct a tree and you can see whether it
distributes over the disjunction ah not so
in this lecture.
What we have done is we have taken into consideration
the semantic gap flux method and then we discussed
in some detail with some examples as for getting
ourselves familiar with this particular kind
of technique so this semantic gap blocks method
is simple to use and the rules are rules are
very few number and then it is easy to use
and it can be implemented in the computer
science field so there are some of the some
important uses for this particular kind of
technique but the problem here is that R.
We are human beings do we use a method like
this particular kind of thing that is a question
that needs to be answered at all is it close
to common sense reasoning or the way we reason
except try and all that there is a difficult
question to answer but as far as implementation
into computers and machines except try not
this technique is going to be a widely used
in so in that context the one which is closer
to the human reasoning is what we call it
as natural deduction method so that is what
we are going to take up in the next lecture
in the next lecture we will be talking about
a natural deduction method in the context
of predicate logic.
