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If p: maths is interesting, q: maths is easy then $p\to \left( \sim p\vee q \right)$ equivalent to:
(a) If maths is easy, then it is interesting.
(b) Math is neither easy, nor interesting.
(c) If maths is interesting, then it is easy.
(d) Maths is neither interesting nor easy.

Answer Verified Verified
Hint: First, solve the Boolean expression to the simplest possible form then interpret it in the form of the statement to reach the answer.

Complete step-by-step answer:

Before moving to the options, let us discuss the meaning of different symbols used in the Boolean expression. See the symbols that we see in the Boolean expressions are called logic symbols, and this includes:

$\wedge $ - represents the AND logical operation.

$\vee $ - represents the OR logical operation.

$\sim $ - represents the NOT logical operator.

$\to $ - represents the if-then logical operator.

Now let us solve the Boolean expression given that is given in the question:

$p\to \left( \sim p\vee q \right)$

We know $x\to y=\sim x\vee y$ , applying this to the above expression, we get:

$\sim p\vee \left( \sim p\vee q \right)$

We also know that the OR logical operation is associative in nature. So, we can change our expression as:

 $\left( \sim p\vee \sim p \right)\vee q$

Now, using the formula: $x\vee x=x$ , we get

$\sim p\vee q$

The above expression is equivalent to $p\to q$ . So, the statement form of the expression becomes “If maths is interesting, then it is easy.” Therefore, the answer is option (c) If maths is interesting, then it is easy.

Note: Be careful about the signs of AND operator and the OR operator, as students generally get confused among their symbols and make mistakes while solving. Also, be careful about the order of the If-then operator, as if-then operator is not commutative in nature. However, the order doesn’t matter in the case of the AND logical operator, and the OR logical operator as they are commutative in nature.