Boolean Algebra

Logic Level 2

$\neg \big(\neg (p \implies q) \implies \neg (q \implies p)\big)$

Which of the given options is equivalent to the above Boolean expression?

\neg p \wedge q

(p\wedge q) \wedge \neg (\neg p \wedge \neg q)

p \iff q

\neg(p \implies q)

1 solution

Ashish Menon
Apr 21, 2016

$\backsim (\backsim (p \rightarrow q) \rightarrow \backsim (q \rightarrow p))$
= $\backsim$ ( $\backsim$ (p $\rightarrow$ q) $\rightarrow$ (q ^ $\backsim$ p))
= $\backsim$ (p $\rightarrow$ q) ^ $\backsim$ (q ^ $\backsim$ p)
= (p ^ $\backsim$ q) ^ ( $\backsim$ q v p)
= (p ^ $\backsim$ q)
= $\boxed{\backsim (p \rightarrow q)}$

Why in step 2 you dont put the negation in the first implication?

Alessandro Rubio - 4 years, 6 months ago

Cuz if i do, then it would become a but more complex and we would have to use the truth tabke to get the final result. Instead lets keep it as it is and bring it in the form $\backsim \left( P \rightarrow Q\right)$ so that we can again simply it into $P \text{^} \backsim Q$ .

Ashish Menon - 4 years, 5 months ago

Boolean Algebra

1 solution

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