Question

If p, q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression \({(p∨q)∧((∼p)∨r)}🡢((∼q)∨r)\) false?

• A. p = F, q = T, r = F
• B. p = T, q = F, r = T
• C. p = T, q = T, r = F
• D. p = T, q = F, r = F

Answer 1

The Correct Option is A

We are given the logical expression: \[ (p \vee q) \land ((\sim p) \vee r) \rightarrow ((\sim q) \vee r). \] We need to find the combination of truth values of \( p, q, r \) that makes the above expression false.

Step 1: Understanding the implication.
The logical expression is an implication:
\[ \text{If} \quad (p \vee q) \land ((\sim p) \vee r) \quad \text{then} \quad (\sim q) \vee r. \] An implication \( A \rightarrow B \) is false only when \( A \) is true and \( B \) is false.
Thus, for the expression to be false, we need to have: \[ (p \vee q) \land ((\sim p) \vee r) = \text{T} \quad \text{and} \quad (\sim q) \vee r = \text{F}. \] 

Step 2: Checking the conditions.
We need to evaluate both parts of the expression for different combinations of \( p, q, r \).
 Case (1) \( p = \text{T}, q = \text{F}, r = \text{T} \):
- \( (p \vee q) = \text{T} \), since \( p = \text{T} \).
- \( (\sim p) \vee r = \text{T} \), since \( r = \text{T} \).
- \( (\sim q) \vee r = \text{T} \), since \( r = \text{T} \).
Thus, the expression is true, not false.
 Case (2) \( p = \text{T}, q = \text{T}, r = \text{F} \):
- \( (p \vee q) = \text{T} \), since \( p = \text{T} \).
- \( (\sim p) \vee r = \text{F} \), since \( p = \text{T} \) and \( r = \text{F} \).
- \( (\sim q) \vee r = \text{F} \), since \( q = \text{T} \) and \( r = \text{F} \).
Thus, the expression is true, not false.
Case (3) \( p = \text{F}, q = \text{T}, r = \text{F} \):
- \( (p \vee q) = \text{T} \), since \( q = \text{T} \).
- \( (\sim p) \vee r = \text{T} \), since \( p = \text{F} \) and \( r = \text{F} \).
- \( (\sim q) \vee r = \text{F} \), since \( q = \text{T} \) and \( r = \text{F} \).
Thus, the expression is false, as \( A = \text{T} \) and \( B = \text{F} \).
 Case (4) \( p = \text{T}, q = \text{F}, r = \text{F} \):
- \( (p \vee q) = \text{T} \), since \( p = \text{T} \).
- \( (\sim p) \vee r = \text{F} \), since \( p = \text{T} \) and \( r = \text{F} \).
- \( (\sim q) \vee r = \text{T} \), since \( q = \text{F} \).
Thus, the expression is true, not false.
 Step 3: Conclusion.
The combination of truth values that makes the expression false is \( p = \text{F}, q = \text{T}, r = \text{F} \), which corresponds to option (1).