and when I solve option (d) , after simplification I get (~a+~b+c) .

So , option (b) is tautology

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Best answer

Answer: (**B**)

$\left(a \wedge b \right) \to b \vee c$

$\implies \neg \left(a \wedge b \right) \vee b \vee c$

$\implies \neg a \vee \neg b \vee b \vee c$

$\implies T$

Option (A) is not TRUE when $c$ is FALSE.

Option (C) is not TRUE when $b$ is TRUE and $c$ is FALSE.

Option (D) is not TRUE when $a$ and $b$ are TRUE and $c$ is FALSE.

-> is right associative according to 2nd ans by http://math.stackexchange.com/questions/12223/associativity-of-logical-connectives

@abhishekmehta4u your approcah is good and all, but a minor mistake in the (A) part :

You accidentally wrote the R.H.S of (A) as** ($b \vee c$ )instead of ($b \wedge c$)**

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