GATE CSE 2014 | Set 1 | Question: 53

Question

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

• A. $(( p  \leftrightarrow q)  \wedge  r)  \vee  (p \wedge q \wedge  \sim r)$
• B. $( \sim (p \leftrightarrow q) \wedge r)\vee (p \wedge q \wedge \sim r)$
• C. $( (p \to q) \wedge r) \vee (p \wedge q \wedge \sim r)$
• D. $(\sim (p \leftrightarrow q) \wedge r) \wedge (p \wedge q \wedge \sim r) $

Comments

Propositional Logic - GATE PYQs Part 2 | Discrete Mathematics | GO Classes | Deepak Poonia (youtube.com)

Timestamp : 1:13:52

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@ritiksri8
Yes, that would be a nice and quick approach to answer this question, but if there were more variables or more cases possible, then the method that @Deepak Poonia sir taught would be much better.

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Exactly 2 can be true among three

so I took P,Q as True , then obviously R becomes False

in this ((p bi-implies q) ^ r) part will become false 

but the q ^ q ^ !r    => this will become true, so the whole equatoin becomes trye 

 

 

and in this part also the ( p ^ q ^ !r) becomes correc -> so the whole equatoin becomes correct

 

 

in this too, ( p ^ q ^ !r) becomes correc -> so the whole equatoin becomes correct

 

 

here both become false

 

 

so why is option A, B and C, three of them not correct ?

 

Answer 1

$A.$ 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"

Comments

https://www.youtube.com/live/7jYCu5n2zkM?si=080T933utW4PSzNE

 

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In option A, won't (p∧q∧∼r) give true if p and q are true, and r is false?

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@GATE 5 now i got it actually in question saying that exactly two of p,q,r should be true.

Actually there will be three cases for exactly two of three should be true:

1)p=T,q=T,r=F   2)p=T,q=F,r=T   3)p=F,q=T,r=T

so in remaining options there is atleast one of these three cases not satisfying but in option B all of these cases are satisfying so option B is correct.

 

Answer 2

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.

Comments

Yes its SOP

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I did the same...

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why not saying how does it satisfy that...All top answers will be so vague...Pls atleast try to make us understand

Answer 3

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.

Answer 4

Correct answer is option B. 

Answer 5

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.

Answer 6

  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

Comments

nice approach