A) \[p\to (p\to q)\]
B) \[p\to (p\,\vee q)\]
C) \[p\to (p\,\wedge q)\]
D) \[p\to (p\leftrightarrow q)\]
Correct Answer: B
Solution :
[b] \[p\to (q\to p)=-p(q\to p)\] \[=\sim p\vee (\sim q\vee p)\] (Since\[p\vee \sim p\]is always true) \[=\sim p\vee p\vee q=p\to (p\vee q)\]
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