Question

Let p,q and r be prepositions and the expression (p$\to$q) $\to$ r be a contradiction. Then, the expression (r$\to$p) $\to$ q is

• A. A tautology
• B. A contradiction
• C. Always TRUE when p is FALSE
• D. Always TRUE when q is TRUE

Answer 1

The correct option is D Always TRUE when q is TRUE

$(p\to q)\to r=0$
i.e.., $(p{}^{\prime }+q){}^{\prime }+r=0$
i.e., $pq{}^{\prime }+r=0$
This is possiable only if $pq{}^{\prime }=0$ and r = 0
Now, $pq{}^{\prime }$ = 0 if p = 0 or $q{}^{\prime }$ = 0
So , final solution is r = 0 and (p = 0 or $q{}^{\prime }=0$
Now let us see
(r $\to p)\to q$
$\equiv (r{}^{\prime }+p){}^{\prime }+q\equiv rp{}^{\prime }+q$
but we have previously shown that r = 0
So, rp${}^{\prime }$ = 0
So, $(r\to p)\to q\equiv rp{}^{\prime }+q\equiv 0+q\equiv q$
So, truth value of (r $\to p)\to$ is same as that of q. So, if is true whenever q is true.
SO option(d) is correct.