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The proposition $\sim$ qvp is equivalent to

  1. p $\rightarrow$ q 
  2. q $\rightarrow$ p 
  3. p $\leftrightarrow$ q 
  4. p $\vee$ q
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$\because (p\lor q)=(\sim p\to q)$

$\therefore (\sim q\lor p)=(\sim(\sim q)\to p)\equiv (q\to p)$

 

we can also verify this with the truth table method:

$p$ $q$ $\sim q$ $(\sim q\lor p)$ $(p\to q)$ $(q\to p)$ $p\leftrightarrow q$ $p\lor q$
T T F T T T T T
T F T T F T F T
F T F F T F F T
F F T T F F T F

From above $(\sim q\lor p) \equiv (q\to p)$

So Option (B) is correct.

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