GO Classes 2024 | Weekly Quiz 7 | Propositional Logic | Question: 1

Answer 1

Solving them By Case Method, Case 1 : p = true, Case 2 : p = false

(A) : Case 1 : (T→ q) V (T → q’) = q V q’ = T Hence a tautology.

        Case 2 : (F → q) V (F → q’) = T V T = T Hence a tautology.

 

(B) : Case 1 : (T → q) V (q → T) = q V T = T. Hence a tautology.

        Case 2 : (F → q) V (q → F) = T V q’ = T. Hence a tautology.

 

(C) : Case 1 : (T → q) V (q → F) = q V q’ = T. Hence a tautology.

        Case 2 : (F → q) V (q → T) = T V T = T. Hence a tautology.

 

(D) : Case 1 : (T → q) V (p → q) = (T → q) V (T → q) = q V q = q. We can stop here as q can either be True or False hence a contingency and not a tautology.

 

Hence, Answer to this question would be option d.

Answer 2

A very good answer is provided by @chokostar already HERE.

For Option A,
$\begin{array}{|c|c||c|c|c|c|} \hline p & q & p\rightarrow q & \neg q & p\rightarrow \neg q &(p\rightarrow q)\vee (p\rightarrow \neg q)\\\hline T & T & T & F& F & T\\\hline T & F & F& T & T & T \\\hline F & T & T & F & T & T\\\hline F & F & T & T & T & T   \\ \hline\end{array}$
The formula is valid since it is satisfied by every interpretation. Similarly, we can check for other options.

Comments

Thankyou sir