in Mathematical Logic edited by
20,964 views
88 votes
88 votes

Which one of the following  well-formed formulae is a tautology?
 

  1. $\forall x \, \exists y \, R(x,y) \, \leftrightarrow \, \exists y \, \forall x \, R(x, y)$
  2. $( \forall x \, [\exists y \, R(x,y) \, \rightarrow \, S(x, y)]) \, \rightarrow \, \forall x \, \exists y \, S(x, y)$
  3. $[ \forall x \, \exists y \, \left( P(x,y) \, \rightarrow \, R(x, y) \right)] \, \leftrightarrow [ \forall x \, \exists y \left(\neg P(x, y) \, \lor R(x, y) \right)]$
  4. $\forall x \, \forall y \, P(x,y) \, \rightarrow \, \forall x \, \forall y \, P(y, x)$
in Mathematical Logic edited by
21.0k views

4 Comments

Ignoring the variables, $(R \rightarrow S) \rightarrow S$ cannot be a tautology.

So, even if we use quantifiers and variables, it cannot be a tautology.
0
0
edited by
[deleted]
0
0

@strawberry-jam I too believe that cause (R->S) -> S

LHS can be ~R + S ->S 

if R = False and S = False

then ~R +S will be True  → S will be false.

 

0
0

5 Answers

67 votes
67 votes
Best answer

NOTE: “Tautology" & “Valid" are Different Concepts in First Order Logic. Tautology & Valid are Same Concepts in Propositional Logic.

Watch the following Detailed Lecture on  “Tautology Vs Valid” in First Order Logic:

$\color{red}{\text{Tautology Vs Valid in First Order Logic:}}$ "Tautology" in Predicate Logic(Click Here)

Note 1 :

When no information is given about Domain of variables for a First Order Logic (FOL) formula, then consider that domain for all variables in the given FOL formula is same. Unless otherwise explicitly stated, domain should be taken same for all variables of a given FOL formula.

Note 2 :

For propositional logic(PL) / Sentential logic, “Tautology” of an expression means same thing as “Validity” of that expression. But this is Not true in FOL.


Tautology/Validity in Propositional Logic :

Tautologies are a key concept in propositional logic, where a tautology is defined as a propositional formula that is true under any possible Boolean valuation of its propositional variables.

Given propositional logic expression (PLE) $G$ is said to be Valid or Tautology iff $G$ is ALWAYS True, regardless of which valuation is used for the propositional variables. 

Some examples of Tautologies in PL :

$A \vee \neg A$

$(A \vee B) \rightarrow (A \vee B)$

$(A \rightarrow B) \rightarrow (\neg B \rightarrow \neg A) $

In propositional logic, there is no distinction between a tautology and a logically valid formula. So, 

In Propositional logic $: \text{Valid} \equiv \text{Tautology} \equiv \text{True}$


Tautology/Validity in First Order Logic(FOL) :

From Wikipedia :

The definition of tautology can be extended to sentences in predicate logic, which may contain quantifiers—a feature absent from sentences of propositional logic. Indeed, in propositional logic, there is no distinction between a tautology and a logically valid formula. In the context of predicate logic, many authors define a tautology to be a sentence that can be obtained by taking a tautology of propositional logic, and uniformly replacing each propositional variable by a first-order formula (one formula per propositional variable). The set of such formulas is a proper subset of the set of logically valid sentences of predicate logic (i.e., sentences that are true in every model).

The fundamental definition of a tautology is in the context of propositional logic. The definition can be extended, however, to sentences in first-order logic. These sentences may contain quantifiers, unlike sentences of propositional logic. In the context of first-order logic, a distinction is maintained between logical validities, sentences that are true in every model, and tautologies, which are a proper subset of the first-order logical validities. In the context of propositional logic, these two terms coincide.

A tautology in first-order logic is a sentence that can be obtained by taking a tautology of propositional logic and uniformly replacing each propositional variable by a first-order formula (one formula per propositional variable). (Note that Propositional logic manipulations can be done on the given formula, but No FOL manipulation should be done.)

For example, because $A \vee \neg A$ is a tautology of propositional logic,  $( \forall x (x=x)  ) \vee  \neg ( \forall x (x=x)  ) $ is a tautology in first order logic. 

Example $2 :$

$[(\forall y \neg Py \rightarrow \neg Px) \rightarrow (Px \rightarrow \neg \forall y \neg Py)]$ is a FOL Tautology.

because it can be obtained from a contraposition tautology  $(A \rightarrow \neg B) \rightarrow (B \rightarrow \neg A) $ by replacing $A$ by $\forall y \neg Py$ and $B$ by $Px$.

Example $3 :$

$ \forall x \forall y P(x,y) \rightarrow \forall x \forall y P(x,y) $ is a FOL tautology because it can be obtained from a PL Tautology $A \rightarrow A $ by replacing $A$ with $\forall x \forall y P(x,y).$

Not all logical validities are tautologies in first-order logic.

For example, the sentence $ {\displaystyle (\forall xRx)\to \lnot \exists x\lnot Rx} $
is true/valid in any first-order interpretation BUT it is Not a FOL Tautology, because it corresponds to the propositional sentence ${\displaystyle A\to B} $  which is not a tautology of propositional logic.

Example $2 :$

$\forall x(Px \rightarrow Px)$  is NOT a FOL Tautology because it corresponds to the propositional sentence $A$  which is not a tautology of propositional logic. Note that $\forall x(Px \rightarrow Px)$ is FOL Valid formula But Not Tautology.

Example $3 :$

$ (\forall x Px) \rightarrow Pc $   is NOT a FOL Tautology because it corresponds to the propositional sentence $A \rightarrow B$ (By replacing $ (\forall x Px)$ with $A$ and $Pc$ with $B$)  which is not a tautology of propositional logic. NOTE that $ (\forall x Px) \rightarrow Pc $ is FOL Valid formula But Not Tautology.

Example $4:$

The following is an example of a logical truth that is not a tautology:  $ \exists x \text{Cube}(x) \vee \exists x \neg \text{Cube}(x) $. Because the validity of the argument depends on the meaning of the existential quantifier $\exists $ , and not just on the meaning of the connective $\vee$.

A sentence containing quantifiers that is a tautology is this: $ \forall x \text{Cube}(x) \vee \neg \forall x \text{Cube}(x) $ which is just an instance of the tautologous form $ A \vee \neg A $.

So that's that. 

NOTE : Every FOL Tautology is FOL validity BUT Not vice versa.

So, If a given FOL formula is Not valid, then it is Not a FOL Tautology. 


Now coming back to the given question :

Option A :

Option A is NOT at all valid in FOL. 

$ \forall x \exists yR(x,y) \leftarrow  \exists y \forall xR(x,y) $ is FOL Validity. But it is Not FOL Tautology.

$ \forall x \exists y R(x,y) \rightarrow   \exists y \forall x R(x,y) $ is NOT FOL Validity. So, it is Not FOL Tautology as well.

Option B :

$ (\forall x[ \exists yR(x,y)\rightarrow S(x,y)])\rightarrow \forall x  \exists y S(x,y) $ is NOT at all valid. 

For example, Let $A$ be a set $\{a,b \}$ and let relation $R$ on $A$ be $  \{ (a,a) \}$  and let relation $S$ on $A$ be $  \{ (a,a) \}$  then clearly  $\forall x \exists y S(x,y) $ is false. Also $ (\forall x[ \exists y R(x,y)\rightarrow S(x,y)]) $ is true. So, we can say that option B is NOT Valid. So, it is Not a Tautology.

Option C :

$ [\forall x \exists y(P(x,y)\rightarrow R(x,y))]\leftrightarrow [\forall x \exists  y(\neg P(x,y)\vee R(x,y))] $  is a FOL Tautology. 

Note that we can do Propositional logic manipulations, so, $A \rightarrow B $ can be written $A' \vee B.$

So, 

$ [\forall x \exists y(P(x,y)\rightarrow R(x,y))] \leftrightarrow [\forall x  \exists y(\neg P(x,y)\vee R(x,y))] $ is same as $ [\forall x \exists y( \neg P(x,y) \vee R(x,y))]\leftrightarrow[\forall x  \exists y(\neg P(x,y)\vee R(x,y))] $

under only propositional logic manipulation.  

Now, we can take $A = [\forall x \exists y( \neg P(x,y) \vee  R(x,y))]$ and so $A  \leftrightarrow A$ is tautology in PL.

Option D :

$ \forall x \forall yP(x,y) \rightarrow  \forall x \forall y P(y,x)$ is FOL Valid But Not FOL Tautology. NOTE that if we replace $\forall x  \forall y P(x,y)$ by $A$ then  $ \forall x  \forall y P(y,x)$ is some B. But $A \rightarrow B$ is Not tautology in PL. 

References :

Page number $115$ in the below book :

Watch the following Detailed Lecture on  “Tautology Vs Valid” in First Order Logic:

$\color{red}{\text{Tautology Vs Valid in First Order Logic:}}$ "Tautology" in Predicate Logic (Click HERE)

edited by

4 Comments

@Deepak Poonia

In option D,

$(\forall x \forall y P(x,y) → \forall x \forall y P(y,x)) ↔ (\forall x \forall y P(x,y) → \forall y \forall x P(y,x)) ↔ (\forall x \forall y P(x,y) → \forall x \forall y P(x,y))$ which turns out to be tautology. Since, quantifiers of the same type are commutative and renaming a variable appropriately won't change the statement.

 

0
0

@Suraj Reddy, See THIS example, it is a Valid expression in FOL, but not tautology.

“Tautology” is a Concept of Propositional Logic, which is only extended to FOL. See THIS.

2
2

@Deepak Poonia 

Sir, In option B in LHS the scope of “There Exists y” should be over R(x,y)  only .
The y present in S(x,y) is bounded by which quantifier?? 

0
0
72 votes
72 votes

A) lets take R(x,y) means x is divisible by y.
X= {6,8,9}
Y={2,3}

For all x, there exists any y such that x is divisible by y.   It is true in our example. 
but here take y individually, no single y is capable of dividing all x.  So it neither implies nor double implies the second one.
So FALSE.

B)  lets take ...
R(x,y)  means  x is divisible by y ,   
S(x,y) means  means   x/y =2 or 3
X={4,6,11}  Y={2,3}
(∀x[∃yR(x,y)→S(x,y)])→∀x∃yS(x,y)

(Always, make the LHS intentionally true, If we can make RHS false, then it is not tautology)

Here for all X, there exists a y such that ..."if x is divisible by y then x/y=2 or 3".... 11 is not divisible by any y...so no problem.
So LHS is true.
Go to RHS.   For all x, there exists a y such that x/y=2 or 3. Which is false. 11/2 and 11/3 are neither 2 nor 3
Here  T-> F  so False. Not tautology.

 D) 
X= 8,12,16
Y= 2,4
For all X and for all Y, Y divides X.......  It does not imply X divides Y.  So FALSE.



We are only left with option C.  So it is the ans. The method I follow is to disprove it. Not to prove it. So here also you can try something by taking any example. You cant disprove it. C is the ans.

edited by
by

4 Comments

Any other approach to solve option B

taking this type of counter-example in exam hall is difficult.
1
1
Very nicely explained!
0
0
nice explanation👍
0
0
45 votes
45 votes

Ans (C).

$\left(P \rightarrow Q\right) \leftrightarrow \left(\neg P \vee Q\right)$

edited by

59 Comments

Yes. But there were two ) missing in C in the GATE paper. ideally they should give marks for All here.
5
5
They are asking for well formed formula C is not well formed formula option D should be correct because its just change of variable names.
1
1
Why can't $\forall x\forall y P(y,x)$ be written as $\forall y\forall x P(x,y)$.

In that case option D will be true
0
0
Because of the priority of ->. You can see the last part of given answer.
1
1
ans C is correct bcoz a<->b meansif  a and b are equivalent then it is tautology ,option has same statement on both sides of<->
1
1
C is correct, but in actual GATE paper the ending braces were missing. GATE takers were supposed to ASSUME it.
0
0
how u assume x<y & is there any other way to solve it not getting this
0
0
P(x, y) is a predicate. And x < y, is just a chosen one, you can chose any other also.

To prove a formula is not valid, it is enough to give one counter example. Here, for D one is given like that.
8
8

The below one is a tautology.

(xyP(x,y))(xyP(y,x)) 

How is this a tautology?

0
0
Because the forall quantifier covers all possible combination of x and y and if it is true, that means there is no combination of x,y for which P(x,y) is false => P(y,x) also cannot be false, as both x and y are coming from same domain.
1
1
edited by

sir, in (d) option how can we think about braces () in exam..as given in both the following statement....

(xyP(x,y))(xyP(y,x)) 

                    or 

x y (P(x,y)  x y P(y,x))

and sir, plz explain more  how the following is tautology??

(xyP(x,y))(xyP(y,x))  

0
0
You should know the precedence rule. $\to$ has higher precedence than quantification.
6
6

(xyP(x,y))(xyP(y,x))  

Here the LHS will be true only if P(x,y) is true for every possible x and y. Since both x and y are coming from the same universe, if this is true, then their order doesn't matter for P(x, y). 

2
2
thanku sir
1
1
Option D is not proved as a contradiction. It is just proved as not tautology.

That is we just gave an instance where D is false. There may or may not be instances where D is TRUE- here there is and hence D is not a contradiction.
1
1
Then FALSE -> FALSE is TRUE.
0
0
How can LHS be true if the relation is not commutative?
0
0
Priority of "->" > "forall or existential"?
0
0
quantification has higher precedence than implication. But a brace, comma or sometimes even "." might be used to change this.
2
2



In the answer, this is given. Why it is evaluated in this way?
Quantification has higher precedence, then why like this?

1
1
yes, so the precedence I told is not obeyed here. I have updated the answer. See now.
1
1
S​ir for D option , is it necessary to consider same domains  for x & y ?
1
1
yes the domain of x and y are not mentioned ,then how can we assume they are from same domain Arjun Sir?
1
1
Yes, I do not think it is right to assume that. So, that is a stronger reason against option D.
2
2
But according to Rosen, Quantification has the highest precedence?? Which one to follow??
1
1
Of course follow Rosen, at least we can argue using Rosen. Wikipedia is not so authentic.
3
3
but why here implication has been given higher precedence??
0
0
Better option among choices :)
0
0

its confusing. i am not getitng it. :-(

(∀xyP(x,y))→(∀xyP(y,x))  is a tautology.

But it wont be true for existential quantifiers right??

0
0
You mean both to $\exists$? Then it should because its just change of variable names.
0
0

yes i mean if the following a tautology??

(∃x∃y)P(x,y))→(∃x∃y)P(y,x)

If yes then y??

0
0
$\exists x \exists y = \exists y \exists x$

As both means there exists x and y.

This proves it rt? Actually implication can even be replaced by double implication.
0
0
edited by
yeah thats completely fine. But the change of order of variable x and y in the predicates doesnt change its meaning?? I am really not able to understand how P(x,y) is same as P(y,x)?? plz clear it.

IF its UNIVERSAL QUANTIFIER then what i understand is that if P(x,y) is true for all  all x and all y, it will be true for all combinations of x and y, So order will not matter if both x and y comes from same domain. But in existential quantifier how its true that order will not matter?? I am sounding silly. lol
0
0
Yes, that can change the meaning. But $\exists x (P(x) ) \leftrightarrow \exists y P(y))$ rt?

Like this, when we put same quantification for both x and y, $P(x,y) \leftrightarrow P(y,x)$. If we change one of them to a different quantifier, then this no longer holds.
3
3
Now i got the meaning. Thanx so much arjun sir. But it will be false if x and y comes from different domain? what is meant by this statemnt?? diffrent domain
1
1
yes, Think of something like every student has a car. So, the domains can be students of a school and say cars inside the parking.
1
1
Thanku so much arjun sir for clearing the concept.
1
1
edited by

x ∀y (P(x,y) → ∀x ∀y P(y,x))

If we give higher precedence to universal quantification and assume same domain for x and y, then this statement  will be true/tautology right?? 

Ideally when quantification has higher precedence than implication why is there a need to assume things?

I know C is a stronger answer but just for knowledge i want the reason.

1
1
Yes, "Why is there a need"

Because GATE question papers are made by humans and not machines. And they do not copy questions. Such such mistakes do come and in such cases options are used to avoid ambiguity. Other way might have been to give "Marks to All", but say you get mark then so do anyone writing GATE which is not really correct as most people would have never even seen Mathematical Logic syllabus. So, justice prevails :)
5
5
Really nice reasoning sir. Marks should not be given to all. Thanx Arjun sir. Learned lot from this question.
1
1
For (∀x[∃yR(x,y)→S(x,y)])→∀x∃yS(x,y), is it correct to say: (∀x∃y[R(x,y)→S(x,y)]) <=> (∀x∃y[R'(x,y) V S(x,y)])?

Please explain how to negate B. Thanks.
0
0
reshown by

Sign <-> represents “not equivalent”.

∀x∃y R( x, y ) is not equivalent to ∃Y ∀X R( X, Y )

Let R( X, Y ) represent X < Y for the set of numbers as the universe, for example. Then ∀X ∃Y R( X, Y ) reads “for every number x, there is a number y that is greater than x”, which is true, while
∃Y ∀X R( X, Y ) reads “there is a number that is greater than every (any) number”, which is not true. So this option is rejected.

Option (d)

Sign -> represents “equivalent”

 ∀X ∀Y R( X, Y ) is equivalent to ∀X ∀Y R( Y, X )

Let R( X, Y ) represent X < Y for the set of numbers as the universe, for example. Then ∀X ∀Y R( X, Y ) reads “for every number X, there is every Y that is greater than x”,
while ∀X ∀Y R( Y, X ) reads “for every number Y, there is every X that is greater than Y”.

And both can’t be equivalent (because at one time, one will be true and other one will be false) So this option is
rejected.

Option (b) is clearly rejected as two predicate can’t be equivalent to one predicate only. So Option (c) is the correct one.

Explanation for option (c) – as position of the quantifier is not changed and since LHS P -> R = ⌐P ᴠ R which is equal to RHS.
Option c is tautology and correct answer too.

1
1

Arjun Sir, 

Is this is a tautology? 

(∃x∃y)P(x,y))→(∃x∃y)P(y,x)

if yes then how. I know you have answered this above but symbols are not showing clearly.

0
0
@hement parihar , it will be true due to there exists, there will be atleast one pair of x,y which will follow this , am i wrong ?
0
0
@hemant

let P(x,y) : x likes y
(∃x∃y)P(x,y) : there exit a x who likes atleast y
(∃x∃y)P(y,x) : there exit a y who liked by atleast x
(∃x∃y)P(x,y))→(∃x∃y)P(y,x)
if (∃x∃y)P(x,y)) is TRUE then (∃x∃y)P(y,x) also be TRUE
1
1
edited by

Precedence Order 

 The syntax for logical connectives is as follows:! (not), ^ (and), v (or), => (implies), <=> (if and only if), FORALL/forall/Forall (universal quantification), and EXIST/exist/Exist (existential quantification). Operator precedence is as follows:

not  > and > or  > implies > if and only if  > forall = exists.

Operators with the same precedence are evaluated left to right. You can use parentheses to enforce a different precedence, or to make precedence explicit (e.g., (A=>B)=>C as opposed to A=>(B=>C)). Universal quantifiers at the outermost level can be omitted, i.e., free variables are interpreted as universally quantified at the outermost level. Quantifiers can be applied to more than one variable at once (e.g., forall x,y). The infix equality sign (e.g., x = y) can be used as a shorthand for the equality predicate (e.g., equals(x,y)).

Link:http://alchemy.cs.washington.edu/user-manual/4_1First_Order_Logic.html

  • But in Kenneth rosen given that the quantifiers Forall (universal quantification) and Exist (existential quantification) have higher precedence than all Logical operators from propositional calculus.

So which one is correct Precedence order? I am just confused.

8
8

@Hemant Parihar

Is this is a tautology? 

(∃x∃y)P(x,y))→(∃x∃y)P(y,x)

Yes.

Actually (∃x∃y)P(x,y) ) === (∃x∃y)P(y,x) means (∃x∃y)P(x,y)) <---> (∃x∃y)P(y,x)

2
2

@chottu Consider this case:

let P(x,y) be "x divides y" and P(y,x) be "y divides x"

Consider x = {2,3} and y = {9}

(∃x∃y)P(x,y)) (LHS) will be true as there exists an x that divides some y. 

(∃x∃y)P(y,x) (RHS) will be false as there doesn't exist any y that divides x. 

This gives T-->F. Where am I going wrong?

2
2

@arjun sir I did not get the answer whether (∃x∃y)P(x,y))→(∃x∃y)P(y,x) is a tautology or not.

Can you please tell again?

0
0

Hi @Warlock lord ji,

Why are you fixing X and Y for both sides ?

means $\exists_{x}\exists_{y}$(P(x,y) $\Leftrightarrow$ P(y,x))  and  $\exists_{x}\exists_{y}$P(x,y) $\Leftrightarrow$ $\exists_{x}\exists_{y}$P(y,x)  are different.

3
3

∀x∀y(P(x,y)→∀x∀yP(y,x))

For every (x,y) if P(x,y) is true then for every (x,y) P(y,x) is true. 

i think this is a mistake in the above answer as the quantifiers have to be applied first i think 

(∀x∀y)(P(x,y))→(∀x∀y)(P(y,x)) is correct

0
0
1
1
#madhab i am nt getting ur statement  "Sign <-> represents “not equivalent”." ?? can anyone explain ??
0
0
∃xα∧β∃xα∧β is (∃xα∧β)(∃xα∧β), and not ∃x(α∧β)∃x(α∧β).
0
0
Option d is not tautology as a relation may or may not be symmetric, therefore this won't be a tautology.
0
0

@, while reading above comment for this point I can't convience myself as this is similar to https://gateoverflow.in/3560/gate2006-it-21 this.

but,one doubt- can $\forall x\forall y$ be distributive over $\rightarrow$.

In Gate 2006 qsn (x,y) from the same relation but here (x,y) may not from the same relation.

0
0
0
0
can you explain a littlle more? how is this a tautology?
0
0
I would immediately mark D and become eligible for a negative mark :(
0
0
0 votes
0 votes

option a :   lhs side means  “for all x there is a y which satisfy r(x,y)”.  and rhs means ”there is a particular y for every x”.  as bi directional implication given this means both LHS and RHS should be equivalent which is false. hence option a is wrong

option b and option d are just random thoughts written in fol .

option c : as we KNOW ,   P(X,Y) →R(X,Y)  ==  ~P(X,Y)  + R(X,Y)  both these are interchangeable and equivalent so option c is right

hence ans c  not so deep explanation  ..but enough for breaking this question and similar ones .this is the way i got ans

 

Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true