 Welcome back in continuation to the last lecture where we discussed about semantic tableaux 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 valid etc. 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 little bit of 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 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 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 and on 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 consistency of these things example if you take into consideration px and qx and then another thing for all x px implies not 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 that always handle the formula which consists of existential quantify so now first you eliminate this quantify using this particular kind of rules suppose if you have a formula like this in your tree then if you remove this existential quantify then you are replacing it with some kind of parameter a and then each time when you remove this existential operator existential quantify you have to use a new parameter where a is new so now if you remove this particular kind of thing then this will become p a and q a so this is one of the instances of this particular kind of formula now we are checking for the consistency of these two formulas now fourth one so now this px implies not qx holds for all x in particular so that is why it holds for even this particular kind of thing one instance of this one could be even this not q a so now we use the same rule x implies y is not x and y so you apply this particular kind of rule for this one then this will become not p a and not q a so now p a and q a can be written in this sense I am just writing it here itself q a 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 have to write like this in the tree diagram so now in this one you have p a here and not p a here this branch closes I mean these two are contradictory to each other a literal and its 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 no q a and not q a here even this branch also closes so that means you list out these statements one after another and you construct a tree diagram and all the branches closes that means there exists some x p x and q x and this particular kind of formula for all x p x implies not q a is said to be inconsistent 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 satisfiable in any one of these interpretations so because all the branches closes so in that sense there exists some x for all x p x implies not q x they are said to be inconsistent to each other so you can replace it with some kind of proposition for p you can replace p with some any other kind of thing in the natural language yourself can see that if you state for all x p x implies not q x and at the same time you say that there exists some x p x and q x then these two statements are to 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 are taking these simple examples then later I move on to check the validity of a given predicate logical form let us say there exists some x p x and q x this is the first statement and you consider another statement such as p x or q x and then third one take any other things such as there exists some x it is not the case that there exists some x p x 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 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 existential quantifier and then one instance of this one is going to be this one now fifth now this can be written as not there exists some x p x 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 so now we can write straight away like this for all x not p x so now here in the second one if you eliminate this existential quantifier 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 this is going to be QB now so now the next one is going to be this one for all x not p x so this is going to be true for all x and all irrespective of whatever you substitute whether a or B it is going to be the case so that is why it is going to be the case not P B so now PA and QA it can be written in this sense QA so that is a fourth one fourth one after another you can write and that second one is P B and QB and then you have not P B once again since you have P B here and not P B here this branch close this it turns out to be the case that these three statements there exists some X P X and not P X and Q X there exists some X P X or Q X sorry this is R we are not supposed to close like this so this is going to be this one P B R Q B so now so this is going to be like this this is P B so this we expanded it and then it will become P B and Q B since there are connected is there which I did not notice it so now not P B and P B closes and then this branch is open so this branch is open in the sense that you have Q A here but you have Q B here so there is no way in which you can cancel you 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 formulas 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 Q B is T and not P B becomes T and then both PA Q A T and this is going to satisfy these three formulas that means they are going to make this 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 need to note that is the list 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 close that 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 or not so now let us talk about some more examples with respect to validity for validity what you will do is for example if you have given a formula X what you need to do is you need to construct a tree diagram for not X and then if all the branches closes that means you land up with a contradiction all the branches closes then not X is considered to be unsatisfiable and that ensures us that X is going to be valid so 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 solves 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 R X Y R Y Z etc and the second statement is for all X for all Y R X Y R Y X and the third statement for all X there exists some Y R X Y and the fourth one for all X R X X so now we want to show whether 1 2 3 the first three statements leads to the fourth statement are not that means fourth 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 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 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 the same because you have to for getting ourselves familiarized 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 Z this is the case R X Y and R Y Z implies R X Z this is the kind of some transitivity property second we have for all X for all Y we have R X Y implies R Y X 3 for all X there exists some Y R fourth one so now this is considered to be the conclusion X 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 you take into consideration the negation of this formula so that is not 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 it is simply written as R X X so just to separate the predicates with the individual variables subscript and superscript and subscript you write in this way so it does not matter 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 or 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 parenthesis 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 and 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 prepositions leads to 4 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 not for all X R X X means this one if you simplify this one and this is there exists some X and then you push this negation inside it will become R X and X so that is what you are going to write here so this is there exists some X not R X so this brackets needs to be clear now so now the strategy for this semantic tableaux method is that first you need to eliminate this existential quantifier before handling the universal quantifies so the best thing to remove best thing to handle is this fifth one so that is when you replace this when you eliminate this existential quantify and we have a rule if something is true this then ? of a where you have to replace this thing with ? of a where a the parameter a 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 617 so now coming back to this one for all X there exists some Y R X Y that means there exists some Y R X Y holds for all X that means it holds for even when you substitute a for that also is going to be it is going to hold for that particular kind of thing also so this is for all X there exists some Y R X Y so now this particular kind of thing holds for all X that means there exists some Y R even if you replace it X with a that is going to hold now so now this is what you are going to write here there exists some Y so how did you get this one five existential instantiation for all Y R so this is a by this is this three universal instantiation because you have removed this universal quantified so now eight since this is the only thing which we have in this one which starts with the existential quantifier we settle with this thing and then we move on to the universal quantifiers so now there exists some Y R a Y 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 one of 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 one so this is going to be R a B rather than R a a because a is already exhausted here so this is seven existential instantiation nine so now coming back to this one so this is over and this is now coming back to this one for all X for all Y R X Y in place R Y Z you take any you substitute any values X and Y that R X Y in place 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 R a B so you substitute X for a and Y for B and this is going to be the case and then R for Y the substituted B and for a substituted a so this is R a B implies R B a so how did you get this one to substitute X for X for a Y for B this is what we have now you expand this one so whenever you have a formula X implies Y it is like this X implies Y is not X and Y so now play this on this one it will become not R a B and then R B a problem is little bit lengthy and all so one 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 is 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 R a B and you have not R a B is exactly contradictory to each other it is like X and not X now this branch closes here itself so now we need to expand this particular kind of branch is this branch which is open there is no BA etc 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 all in the case of preposition logic for example if your P ? Q ? Q etc and all R ? B while constructing the semantic tab blocks 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 X Q X this can be used n number of times because really you can use the same form because it happens for all X so in the case of preposition logic each time when you are expanding the tree with this formulas you are checking this formulas next time when you do it when you check this particular find a formula you do like this not R not R P not R so now you check this formula and all the formulas are checked so you put tick mark for this particular kind of thing that is not going to happen in this particular kind of situation it can be used recursively and all 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 Y for B and Z for C Z for E so wherever you have X you substitute with A and wherever you have Y you substitute with B and wherever you have Z you substitute again with the A that substitution should be uniform then this formula will become example here you have like this for all X for all Y for all Z R X Y and R Y Z implies R X Z so now you are substituted like this X for A wherever you have X you substitute with A wherever you have Y substitute with B and then wherever you have Z you substitute with A so now this will become R now you are eliminating this quantifiers now one instance of that one is this particular kind of thing so now this will become a B the first one and R Y Z means instead of Y we have B here B and A implies this one R X Z or X means A Z means X and Z are same so that is why we did like this because we have a term R BA somehow you 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 R so now if you further simplify this one so this is X implies Y so now this will become not of not of R so this is not X not R A B and R BA and then this simplifies to this one R A and A so now this is going to be like this not of R AB is like this not of R A and B and this is not of R B and A 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 just remove this particular kind of thing 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 and you have not R A B and then you have not of R AA and then all the way down these are the things which we have so now observe this particular kind of thing not of R AA and you have not R AA is a branch close actually this should be like this since we do not have space here so we have gone the other way round so not R AA and then you have R AA this closes now coming back to this one R AB and not R AB this branch closes now not R BA there is something called not R BA is this here R BA is there here and then all the way down here you have to write this also this I forgotten R BA is there and not R BA is there even this also closes 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 valid valid means it has to be true that means this is the this is considered to be the the original conclusion is considered to be the true kind of conclusion that means this follows from these three statements so in the same way you can check whether 1 and 2 leads to 3 2 and 3 leads to 1 all these things you can check just 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 unsatisfiability that means negation of X is considered to be contradiction that means X has to be the case X has to be true X has to be tautology so 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 are not 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 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 a semantic tableaux method this one or two examples we will be considering and then we will end this lecture so now let us consider it is coming back to the consistency again the problem of consistency let us say we have a set like this Px implies Qx and then there exists some x Px and there exists some x not Qx so we have the in then in our set we have these three formulas so now we are checking whether these three formulas are consistent to each other are not so now you start you numbering those things one two three now we are checking the consistency so the first thing that you do is as usual in the semantic tableaux method in the predicate logic is that you have to handle the predict existential quantifiers first you can handle any one of these things now if you eliminate this existential quantifier here is an instance PA to existential instantiation 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 a it has to be B so this is three existential instantiation is this one so now 61 Px ? Qx holds for all x so that is why it has to hold for PA ? QA it has to be true for even PB ? QB you can you can all you can also use that particular kind of things so we have used this PA in place QA now this is going to be PA QA so PA and not a closes and not QA and you have not QB and this branch opens that means this particular kind of interpretation satisfies this three formulas and all that makes this three formulas true together that means this these three statements are said to be consistent to each other 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 in a different so now let us see whether you take these two statements into consideration now whether or not this not Qx follows from these two statements are not so now so now we write it in the conclusion now so for all x Px ? Qx there exists some x Px and then there exists some x not Qx whether this follows are not from these two premises so how do we check whether or not this argument is valid or not so again we use the semantic tableaux method in that the first step includes the negation of the conclusion that is not there exists some x not Qx you start with this particular kind of thing so now here we have a definition for all x Qx is same as there does not exist some x not x in the same way there exists some x Qx is same as not for all x not x this is the standard definitions and all so exist universal quantifier can be defined in terms of existential quantifier than existential quantifier 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 Qx 3 by definition so what we have done here taken 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 are 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 so to instantiation is this one PA so now six one you can handle any one of these things now they all these two starts with for all x of the so now one instance of this one is going to be PA sorry this is going to be not PA implies QA so this is the instance of this one so universal instantiation of this one is going to be PA implies QA so this is if you expand this one it is going to be lies not PA and QA now you have another formula this thing for all x Qx that means it has to be true for even a also so that is why you can write it straight away like this 4.1 Q A so this is for universal instantiation this so now PA and not PA but this branch remains open that means from these two premises that means 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 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 and the conclusion is false and that is considered to be any count it considered to be a counter example for the given argument and hence that argument is invalid so here is an instance where you are even if you deny the conclusion you still have it still makes this satisfiable and all that means true premises and false conclusion is going to be satisfiable in this particular kind of thing especially when Q A is true PA is true then this whole statements are going to be true that means your true premises and a false conclusion that so 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 rivers or anything and then in that particular kind of thing you need to have some kind of relation or in particular predicate and then you whatever is true here you listed out Q A and PA are true and based on that you can judge that you know I can easily construct a counter example for 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 invalid so in this way we can solve some difficult problems as well this consider one more example and then we will finish it off so in the context of distribution of universal quantifiers we asked ourselves whether universal quantifiers are distributed or not so that is a particular kind of thing for example if we have this formula for all X P X R Q X so from this whether or not it follows that means whether we can derive this particular kind of for all X P X or for all X Q X if for all X P X implies for all X P X are for all X way then this is distributed over this distributed over disjunction universal quantifiers are distributed over the disjunction so again we want to see whether this particular kind of thing holds are not so P X R Q X and then here it is individually we have written like this so now again the using the semantic tableau method whether this argument this follows are not is the one which we are trying to check so you list out the premises like this P X R Q X and then you start with the negation of the conclusion for all X P X for all X Q X so now three you simplify this particular kind of thing then it will become for all X P X negation of the disjunction will become conjunction and then this will be for all X Q X so now this can be written in this sense for all X P X 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 you get this one so now fourth six one now you further simplify this one not for all X P X not for all X P X the same as there exists some X you put this negation inside and then it will be it is not the case so there exists some X not P X and then seventh one there exists some X not Q X so how did we get this one for by definition and five by definition the definition is this one so the problem is not it over now so now we have there exists some X not P X there exists some X not 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 be a so this is six existential quantization and then eight seven if you apply extension again then this will be not Q it should not be a it should be B now we have this for all X P X implies a cure that is going to hold for any any value any parameter you can substitute into it that is going to hold that means it satisfies that particular kind of formula so now we have not PA and we have not Q B and then for example if you substitute it as PA and so one instance of this one is going to be PA or Q A it can very well be like this also PB or Q B also so we take this into consideration then this will be like this PA 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 Q A you have taken into consideration PB and Q B so then what will happen 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 X P X R Q X you will not be able to derive for all X P X R for all X Q X so now that means that it does not imply this particular kind of thing you can check whether you saw 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 are now so in this lecture what we have done is we have taken into consideration the semantic tablax method and then we discussed in some detail with some examples as for getting ourselves familiarized with this particular kind of technique so this semantic tablax method is simple to use and the rules are rules are very few in number and then it is easy to use and it can be implemented in computers as well so there are some of the some important uses for this particular kind of technique but the problem here is that are we are human beings do we use a method like this particular kind of thing that is a question that needs to be answered and all is it close to common sense reasoning or the way we reason etc and all that is a difficult question to answer but as far as the implementation into computers and machines etc and all 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 the natural deduction method in the context of predicate logic.