GO Classes 2023 | Weekly Quiz 3 | Question: 14

Answer 1

Answer: C, D

1.  $F1\wedge F2\wedge F3$ is unsatisfiable. this means they are always false, i.e. contradiction.
2. Conjunction of any pair of them is satisfiable, i.e. at least one time true.
   1. $F1\wedge F2$ will be true in one case, say k1. Then, F1=T and F2=T in this case.
   2. $F2\wedge F3$ will be true in one case, say k2. Then, F2=T and F3=T in this case.
   3. $F1\wedge F3$ will be true in one case, say k3. Then, F1=T and F3=T in this case.

Lets check the options one by one.

• A. At least one of F1,F2,F3 is a tautology.

• Assume F1 is tautology. So, F1 will be always True. So,  for case k2, when F2=T and F3=T. $F1\wedge F2\wedge F3$ will be true. This contradicts (i). Similarly, we can prove for F2 and F3. So this option is False.

• A. At least one of F1,F2,F3 is a contradiction.

• Assume F1 is Contradiction. So, F1 will be always False. So, now for case k1 and k3, $F1\wedge F2$, and $F1\wedge F3$ will be False. This contradicts (ii)- 1. and 2. Similarly, we can prove for F2 and F3. So, this option is False.

100. F1,F2,F3, all are contingency.

• From option A. and B. , we proved that F, F2, F3 are neither Tautology nor Contradiction. So, definitely they are contingency. So, this option is True.

500. (F1→F2),(F2→F3),(F3→F1), all are contingency.

• Consider (F1→F2), for case k1, F1=T and F2=T. So, (F1→F2) is True in this case. Now, for case k3, F1=T and F3=T. But, F2 will be False in this case, because of (i). so, as F1=T and F2=F, so (F1→F2) will be False. So, as (F1→F2) is False in one case and True in one case. So, it is contingency. Similarly, we can prove for others. so, this option is also True.