Mathematics | Predicates and Quantifiers | Set 2

Prerequisite : Predicates and Quantifiers Set 1, Propositional Equivalences

Logical Equivalences involving Quantifiers
Two logical statements involving predicates and quantifiers are considered equivalent if and only if they have the same truth value no matter which predicates are substituted into these statements irrespective of the domain used for the variables in the propositions.
There are two very important equivalences involving quantifiers, given below-

1.  forall x(P(x)wedge Q(x)) equiv forall xP(x) wedge forall xQ(x)

2.  exists x(P(x)vee Q(x)) equiv exists xP(x) vee exists xQ(x) 

One is compelled to think whether the equivalences would hold if the conjunction is replaced with disjunction in (1) and disjunction is replaced with the conjunction in (2). The answer may seem like Yes, but on second thought, you would realize that the answer is No.

To prove why they are not equivalent, we must understand what makes two statements equivalent. As explained in the previous article Propositional Equivalences two statements P and Q are equivalent if-

PLeftrightarrow Q, which can also be restated as PRightarrow Q wedge QRightarrow P.

If they are equivalent then,
forall x(P(x)vee Q(x)) Leftrightarrow forall xP(x) vee forall xQ(x) and,
exists x(P(x)wedge Q(x)) Leftrightarrow exists xP(x) wedge exists xQ(x)
both must be true.

Let us first check for forall x(P(x)vee Q(x)) Leftrightarrow forall xP(x) vee forall xQ(x).
Is forall x(P(x)vee Q(x)) Rightarrow forall xP(x) vee forall xQ(x) true?

Proof – Suppose that the Hypothesis forall x(P(x)vee Q(x)) is true. That means there are certain x for which P(x) is true and others where Q(x) is true.
It is also possible that for some x both P(x) and Q(x) are true. But in any case, all x must either satisfy P(x) or Q(x) or both, since the hypothesis is true.
The conclusion(RHS) is true when the disjunction is true. As is clear from the above reasoning that P(x) is true for some values of x and Q(x) for some.
Thus both forall xP(x) and forall xQ(x) are false, since neither of them are true for all values of x.
In the case where P(x) and Q(x) hold for all x then this equivalence is true, but otherwise it is false.

So, forall xP(x) vee forall xQ(x) equiv F. According to our assumption, the hypothesis is true, but our conclusion turned out to be false. This cannot be true for a conditional, therefore the conditional
forall x(P(x)vee Q(x)) Rightarrow forall xP(x) vee forall xQ(x) is false.

Since one conditional is false, the complete biconditional is false.
Hence, forall x(P(x)vee Q(x)) 
otequiv forall xP(x) vee forall xQ(x).

In a similar way, it can also be proved that,
exists x(P(x)wedge Q(x)) 
otequiv exists xP(x) wedge exists xQ(x)

As an exercise prove the above non-equivalence and also the equivalences involving quantifiers stated above. Remember to prove the bi-conditional and not just one conditional.

Negating Quantified statements
Consider the statement “Every Computer Science Graduate has taken a course in Discrete Mathematics.”
The above statement is a universal quantification, xP(x)
where P(x) is the statement “x has taken a course in Discrete Mathematics” and the domain of x is all Computer Science Graduates.
The negation of this statement is “It is not the case that every computer science graduate has taken a course in Discrete Mathematics” or simply “There is a computer science graduate who has not taken a course in Discrete Mathematics”.
The above statement can be expressed using an existential quantification.
exists x 
eg P(x)
Thus, we get the following logical equivalence-

eg forall xP(x) equiv exists x 
eg P(x)

eg exists xP(x) equiv forall x 
eg P(x)
These equivalences are nothing but rules for negations of quantifiers. They are also known as De Morgans’s laws for quantifiers.

In summary,
 egin{tabular}{||c||c||c||c||} hline Negation & Equivalent statement & When true? & When false?\ hline hline 
eg exists xP(x) & forall x 
eg P(x) & P(x) equiv F,:for:all:x & P(x) equiv T, :for:some:x \ hline 
eg forall xP(x) & exists x 
eg P(x) & 
eg P(x) equiv T,:for:some:x & P(x) equiv T, :for:all:x \ hline end{tabular} 

Nested Quantifiers
It is possible to use two quantifiers such that one quantifier is within the scope of the other one. In such cases the quantifiers are said to be nested.
For example, forall x exists y (x + y = 0)
The above statement is read as “For all x, there exists a y such that x +  y = 0.

Note: The relative order in which the quantifiers are placed is important unless all the quantifiers are of the same kind i.e. all are universal quantifiers or all are existential quantifiers.

In summary,
 egin{tabular}{||c||c||c||} hline Statement & When true? & When False? \ hline hline forall x forall y P(x,y) & P(x,y) equiv T: for: every: (x,:y)& P(x,y) equiv F : for :some:(x,:y)\ forall y forall x P(x,y) & & \ hline   forall x exists y P(x,y) &  shortstack{For:every:x:there:is:a:y:such:that \ P(x,y) equiv T}&shortstack{There:is:an:x:such:that\P(x,y) equiv F : for :all:y}\[3ex] hline  exists x forall y P(x,y) &shortstack{There:is:an:x:such:that\P(x,y) equiv T : for :all:y}&  shortstack{For:every:x:there:is:a:y:such:that\P(x,y) equiv F:}\[3ex] hline  exists x exists y P(x,y) & P(x,y) equiv T: for: some: (x,:y)& P(x,y) equiv F : for :all:(x,:y)\ exists y exists x P(x,y) & & \ hline end{tabular} 

GATE CS Corner Questions

Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or in GATE Mock Tests. It is highly recommended that you practice them.
1. GATE CS 2012, Question 17
2. GATE CS 2013, Question 27
3. GATE CS 2013, Question 47
4. GATE CS 2010, Question 30
5. GATE CS 2009, Question 26
6. GATE CS 2005, Question 36
7. GATE CS 2016 Set-2, Question 37
Majority of the questions asked in GATE from Discrete Mathematics focus on Predicate Logic. Almost all of them involve quantifiers.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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