GATE CSE 2017 | Set 1 | Question: 29

Question

Let $p$, $q$ and $r$ be propositions and the expression $\left ( p\rightarrow q \right )\rightarrow r$ be a contradiction. Then, the expression $\left ( r\rightarrow p \right )\rightarrow q$ is

• A. a tautology
• B. a contradiction
• C. always TRUE when $p$ is FALSE
• D. always TRUE when $q$ is TRUE

Comments

if we find the values from contradiction statement then P=Q=True and R is false.

if we expand second statement then we get (r AND p) OR q. Now as we have calculated values using contradiction it will always be true as Q is true. if we solve just second without using previously calculated values then it depends on Q.

 

So my question is the values that we get from contradiction are to be used in proving second statement yes OR no?

Thanks in advace.

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$((p \rightarrow q) \rightarrow r) = F, \text{   } \therefore (r=F)$

$\therefore (r \rightarrow p) \rightarrow q = (r’ + p)’ + q = (T + p)’ + q = q$

It's always faster to solve since the first part can be intuitively calculated.

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on solving the first eqn we get 3 values of p, q and r.

we place them on the 2nd eqn and check on 4 eqns, option d is result.

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Option D is correct

Answer 1

Given $(p\to q)\to r$ is false. It is possible only when $r$ is FALSE and $(p\to q)$ is TRUE.
Now even without checking any other option we can directly conclude option $D$ is correct as $(r\to p)\to q$ can be written as $\neg(r\to p)\vee q.$

Since, $r$ is False, $r\to p$ is true and $\neg (r\to p)$ is False. So, it becomes $(\text{False} \vee q).$
which is TRUE whenever $q$ is TRUE

Hence, option (D)

Comments

As (P→Q)→R is false. It is possible only when R is FALSE and (P→Q) is TRUE.

But (P→Q) is true only when Q is True and P can be anything True or False but  if Q become false and P become True then (P→Q) is False then here (P→Q) is not True so (P→Q)→R give True as (P→Q) and R both are False so (P→Q)→R is not remain contradiction.

so to remain contradiction Q should be True and R should be False.

(R→P)→Q is Always True as (False→True/False)→True=True→True= True

so answer should be option (A).

Please let me know if I am wrong?

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$\left ( P\rightarrow Q \right )\rightarrow R := P\wedge \sim Q \vee R$  Right ? @Arjun Sir Can you please check ?

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(P→Q)→R   =   (P∧∼Q)∨R

@ghostman23111

Answer 2

D)

Given that ( p → q ) → r  is a contradiction so  r = F and ( p → q ) = T 

We have to evaluate the expression ( r → p ) → q  

Since r = F, ( r → p )  = T ( As F → p, is always true )

The final expression is T → q and this is True when q is True, hence option D.  

Comments

Yes always true for q true.

Answer 3

so ans D is correct.

Answer 4

Solution.....

Answer 5

If a given compound proposition is a contradiction then it is FALSE for every possible combination of the truth value of  it’s atomic propositions

Answer 6

Given, ((p → q) → r) $\equiv$ F  => (p → q) $\equiv$ T and r $\equiv$ F  (T: true ; F: false)

((r → p) → q) $\equiv$ ((F → p) → q) $\equiv$ (T → q) $\equiv$ q

$\therefore$ The expression ((r → p) → q) $\equiv$ q is a contingency. It is always true when q is true.