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

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37 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) $

41 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"

I feel the answer is right but explanation for option (A) is wrong in selected answer.

Just think what does ”q if p” sentence format say in propositional logic. It means **p->q**. If you see it’s truth table,it means when ‘p’ is true , ‘q’ should be compulsorily TRUE.But when ‘p’ is false , ‘q’ can assume any value.

→ Now here ‘p’=”EXACTLY TWO OF p,q,r are TRUE.”

→ ‘q’ = “Given option is right”.

(A) is wrong bcoz when ‘q’ and ‘r’ are TRUE and ‘p’ is false , it assumes FALSE value, which shouldn’t happen as discussed above.

(B) is right .

(C) and (D) are false for same reason as (A)

1

26 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.

18 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.**

3 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.