Question

Which of the following is /are tautology:

• A. $(a\vee b)\to (b\wedge c)$
• B. $(a\wedge b)\to (b\vee c)$
• C. $(a\vee b)\to (b\to c)$
• D. $(a\to b)\to (b\to c)$

Answer 1

The correct option is B $(a\wedge b)\to (b\vee c)$

(a)
$(a\vee b)\to (b\wedge c)$
$\equiv (a+b){}^{\prime }+bc$
$\equiv a{}^{\prime }b{}^{\prime }$ +bc
Therefore, $((a\vee b)\to (b\wedge c))$ is contingency and not tautology.

(b)
$(a\wedge b)\to (b\vee c)$
$\equiv ab\to b+a$
$\equiv \left(ab\right){}^{\prime }+b+c$
$\equiv a{}^{\prime }+b{}^{\prime }+b+c$
$\equiv a{}^{\prime }+1+c$
$\equiv 1$
So $((a\wedge b)\to (b\vee c))$ is tautology.

(c)
$(a\vee b)\to (b\to c)$
$\equiv (a+b)\to (b{}^{\prime }+c)$
$\equiv (a+b){}^{\prime }+b{}^{\prime }+c$
$\equiv a{}^{\prime }+b{}^{\prime }+b{}^{\prime }+c$
$\equiv b{}^{\prime }+c$

So $((a\vee b)\to (b\to c))$ is contingency but not tautology.

(d)
$((a\to b)\to (b\to c))$
$\equiv (a{}^{\prime }+b)\to (b{}^{\prime }+c)$
$\equiv (a{}^{\prime }+b){}^{\prime }+b{}^{\prime }+c$
$\equiv ab{}^{\prime }+b{}^{\prime }+c$
$\equiv b{}^{\prime }+c$

Therefore,$((a\to b)\to (b\to c))$ is contingency but not tautology.