**B is correct ****one !**** Here is better clarification:**

29 votes

Which one of the following propositional logic formulas is TRUE when exactly two of $p,q$ and $r$ are TRUE?

- $(( p \leftrightarrow q) \wedge r) \vee (p \wedge q \wedge \sim r)$
- $( \sim (p \leftrightarrow q) \wedge r)\vee (p \wedge q \wedge \sim r)$
- $( (p \to q) \wedge r) \vee (p \wedge q \wedge \sim r)$
- $(\sim (p \leftrightarrow q) \wedge r) \wedge (p \wedge q \wedge \sim r) $

33 votes

Best answer

$A.$ will be true if $P,Q,R$ are true, $((p ↔q) ∧ r)$ will return true. So "exactly two" is false

$C.$ if only $r$ is true and $p$ and $q$ are false, first part of implication itself will result in true

$D.$ if $r$ is true or false, this returns false due to $r$ and $\neg r$ present in conjunction. So, this is a CONTRADICTION.

B is the answer. B is true if $p$ is TRUE and $q$ is FALSE or vice verse, and $r$ is true or if $p$ and $q$ are TRUE and $r$ is FALSE.

PS: Actually the question should have been "TRUE **ONLY** when exactly two of $p,q$ and $r$ are TRUE"

2

**Why A is false.** If P=Q=T and R = F then option A, B and C becomes true.

because (P^Q^~R) will be true and as it is in OR-ing then whole expression becomes true.

0

Do we need to modify the explanation Arjun sir???? Could you please check the video link that I posted below?

1

getting a,b,c as answer

A is true when p-T,q-T,r-F

B is true when p-F,q-T,r-T

C is true when p-F,q-T,r-T

Can you please explain why only B is selected ?

17

Actually the question says "exactly two". So, we must also ensure that the formula is never TRUE if all the variables are (or if only 1 variable is) TRUE and also ensure that whenever exactly two of the variables are TRUE, it must evaluate to TRUE.

2

To get answer as B is the question must be Which one of the following propositional logic formulas is TRUE **ONLY** when exactly two of *p*,*q* and *r** *are **TRUE**?

3

Grammatically you are right :) But sometimes we need to assume stuffs especially for last 2 years of GATE where they stopped giving Marks to All for debatable questions.

0

sir ,

please explain little bit more...how do you take exactly 2 , i mean it can be p,q or q,r or r,p .. should we check for all cases in all option ...? actually how to solve it ...

please explain little bit more...how do you take exactly 2 , i mean it can be p,q or q,r or r,p .. should we check for all cases in all option ...? actually how to solve it ...

16 votes

the proposition logic formula is required to be true when any two of the p q r are true .so let's make truth table taking any two of p q r as true

P Q R

0 1 1

1 1 0

1 0 1

now write down the POS for above truth table

i.e. P'QR+PQR'+PQ'R .

option B will satisfy the above POS expression hence answer.

P Q R

0 1 1

1 1 0

1 0 1

now write down the POS for above truth table

i.e. P'QR+PQR'+PQ'R .

option B will satisfy the above POS expression hence answer.

12 votes

propositional logic formulas is TRUE when exactly two of p,q and r are TRUE means possibilities are pq or qr or pr to be TRUE

**option A)**

((p↔q)∧r)∨(p∧q∧∼r)

=((T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=((T)∧r)∨(T∧∼r)

=(r)∨(∼r)

=T

It doesn't depend upon the value of r, so, if the value of r is T then it also to be T so the condition exactly 2 are T is false. Hence **option A is FALSE.**

**option B)**

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=(∼(T)∧r)∨(T∧∼r)

=(F∧r)∨(T∧∼r)

=(F)∨(∼r)

=∼r

so if r=F then this is T so condition exactly 2 are T is true

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(T↔q)∧T)∨(T∧q∧∼T) ( take p=T and r=T)

=(∼(q)∧T)∨(F)

=∼(q)∨(F)

=∼q

so if q=F then this is T so condition exactly 2 are T is true

(∼(p↔q)∧r)∨(p∧q∧∼r)

=(∼(p↔T)∧T)∨(p∧T∧∼T) ( take q=T and r=T)

=(∼(p)∧T)∨(F)

=∼(p)∨(F)

=∼p

so if p=F then this is T so condition exactly 2 are T is true

**Hence option B is CORRECT.**

**option C)**

((p→q)∧r)∨(p∧q∧∼r)

=((T→T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)

=((T)∧r)∨(T∧∼r)

=(r)∨(∼r)

=T

It doesnt depend upon value of r , so , if value of r is T then it also to be T so the condition exactly 2 are T is false . **Hence option C is FALSE.**

**option D)**

(∼(p↔q)∧r)∧(p∧q∧∼r)

=(∼(T↔T)∧r)∧(T∧T∧∼r) ( take q=T and r=T)

=(∼(T)∧r)∧(T∧∼r)

=(F∧r)∧(T∧∼r)

=F∧(∼r)

=F

It doesnt depend upon value of r , so , if value of r is F then it also to be F so the condition exactly 2 are T is false . **Hence option D is FALSE.**

2 votes

Answer is (B) here any two propositional variables can be true and any one of p,q and r can be false .So let's start with elimination:

D is eliminated as D won't be true if p ,q will be true and r is false since there is an and between two expressions.

In C if q is false and p and r is true then the expression:(~p or q)AND r becomes false which doesn't satisfy the condition .

Coming to B it seems to be the right answer even when p is false ,q and r is true and when q is false and p and r is true.

A won't satisfy the condition if anyone p or q is false.

The key is : p<->q is actually (p->q)AND(q->p).

P->q is ~p or q.you just need to substitute T and F and check for condition.

D is eliminated as D won't be true if p ,q will be true and r is false since there is an and between two expressions.

In C if q is false and p and r is true then the expression:(~p or q)AND r becomes false which doesn't satisfy the condition .

Coming to B it seems to be the right answer even when p is false ,q and r is true and when q is false and p and r is true.

A won't satisfy the condition if anyone p or q is false.

The key is : p<->q is actually (p->q)AND(q->p).

P->q is ~p or q.you just need to substitute T and F and check for condition.

0 votes

We assume ‘p’ and ‘q’ to be TRUE. Therefore, ‘r’ is false because exactly two of ‘p’, ‘q’, and ‘r’ can be TRUE.

**p <--> q** means if both ‘p’ and ‘q’ have same values then **p <--> q** is TRUE. So, **((p <--> r ) and r)** evaluates to be FALSE. Therefore, **~((p <--> r ) and r)** is TRUE.

Here, two sub- expression are connected via OR operator and one of the sub-expression is TRUE. So, the complete expression becomes TRUE.

Therefore, option (B) evaluates to be TRUE.

0 votes

If you check for 110

A,B, C all are true.

If you check for 111,

A, C are true, while B is false.

So, B is the answer.

A,B, C all are true.

If you check for 111,

A, C are true, while B is false.

So, B is the answer.

0 votes

Consider 4,6,7 cases and check options.

$B)\\ (\sim (p\leftrightarrow q)\wedge r) \vee (p \wedge q \wedge \sim r)\\$

$4^{th}$ case

$=(\sim (F\leftrightarrow T)\wedge T) \vee (F \wedge T \wedge \sim T)\\ =(\sim F\wedge T) \vee F\\ =(T\wedge T) \vee F\\ =T \vee F\\ =T$

Similarly Check $6^{th} \&\ 7^{th}$ cases.

**Correct Answer: B**