Question

$( S 1)(p \Rightarrow q) \vee(p \wedge(\sim q))$ is a tautology(S2) $((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$ is a contradictionThen

• A. both (S1) and (S2) are correct
• B. only (S2) is correct
• C. only ( $S 1)$ is correct
• D. both ( $S 1)$ and (S2) are wrong

Hint:

To check whether a logical expression is a tautology or contradiction, use truth tables. A tautology is true for all possible truth values, while a contradiction is false for all possible truth values.

Answer 1

The Correct Option is C

p	q	p ⇒ q	∼q	p∧∼q	(p ⇒ q) ∨ (p∧∼q)
T	T	T	F	F	T
T	F	F	T	T	T
F	T	T	F	F	T
F	F	T	T	F	T

 

∼p	∼q	∼p ⇒ ∼q	∼p ∨ q	((∼p) ⇒ (∼q)) ∧ (∼p)∨q)
F	F	T	T	T
F	T	T	F	F
T	F	F	T	F
T	T	T	T	T

Answer 2

Step 1: Analyze \( S1 \): \[ S1 = (p \Rightarrow q) \vee (p \land \neg q). \] Let's check its truth table: \[ \begin{array}{|c|c|c|c|} \hline p & q & p \Rightarrow q & p \land \neg q \\ \hline T & T & T & F \\ T & F & F & T \\ F & T & T & F \\ F & F & T & F \\ \hline \end{array} \] For \( (p \Rightarrow q) \vee (p \land \neg q) \), we evaluate the disjunction in each case: 
- When \( p = T \) and \( q = T \), \( (p \Rightarrow q) = T \), and \( (p \land \neg q) = F \), so the result is \( T \). 
- When \( p = T \) and \( q = F \), \( (p \Rightarrow q) = F \), and \( (p \land \neg q) = T \), so the result is \( T \). 
- When \( p = F \) and \( q = T \), \( (p \Rightarrow q) = T \), and \( (p \land \neg q) = F \), so the result is \( T \). 
- When \( p = F \) and \( q = F \), \( (p \Rightarrow q) = T \), and \( (p \land \neg q) = F \), so the result is \( T \). 
Thus, it evaluates to true in all cases, which means \( S1 \) is a tautology. 
Step 2: Analyze \( S2 \): \[ S2 = (\neg p) \Rightarrow (\neg q) \land ((\neg p) \vee q). \] Let's check its truth table: \[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \neg p & \neg q & (\neg p) \Rightarrow (\neg q) & (\neg p) \vee q \\ \hline T & T & F & F & T & T \\ T & F & F & T & T & F \\ F & T & T & F & F & T \\ F & F & T & T & T & T \\ \hline \end{array} \] From the truth table, we see that \( S2 \) does not evaluate to false for all cases, meaning \( S2 \) is not a contradiction. It evaluates to true in some cases, and therefore \( S2 \) is not incorrect but has cases where it is true.