A) \[p\vee q\]
B) \[p\Rightarrow q\]
C) \[p\wedge (\sim q)\]
D) \[p\Rightarrow p\]
Correct Answer: B
Solution :
When p is true and q is false, then \[p\vee q\] is true, \[q\Rightarrow p\] is true and \[p\wedge (\sim q)\] true. (\[\because \] both p and \[\sim q\] are true) Here, p => q is not true as a true statement cannot imply a false statement.
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