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.

---

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

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

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

Answer 1

’

I found this method easy. 

Answer 2

.