1. ## [SOLVED] Boolean Simplification

Hi guys,

I've tried to simplify a formula, and I'm not sure if I've done it right. The following is my solution - I've indicated which logic rule(s) I've used to arrive at each line (excuse the rubbish handwriting):

http://www.zaraphrax.net/~fagg/boolean_mhf.jpg

Any guidance is appreciated.

Thanks!

2. Originally Posted by Ashgasm
Hi guys,

I've tried to simplify a formula, and I'm not sure if I've done it right. The following is my solution - I've indicated which logic rule(s) I've used to arrive at each line (excuse the rubbish handwriting):

http://www.zaraphrax.net/~fagg/boolean_mhf.jpg

Any guidance is appreciated.

Thanks!
I think that your solution is ok. You may notice that the last line is in fact $\displaystyle p\Rightarrow q$.

You can also use truth table to verify that your answer is correct.
For proposition which have the form of implication there is a shorter way:
Clearly $\displaystyle p\Rightarrow q$ if false only in one case: p=1, q=0.
Let us try to the same thing with the first expression
$\displaystyle [p\land \neg(p\Rightarrow q)] \Rightarrow q$.
This implication is false only if $\displaystyle [p\land \neg(p\Rightarrow q)]=1$ and q=0. For a conjunction to be true, you need that both of the operands are true. This means that p=1, and $\displaystyle \neg(p\Rightarrow q)$ is also true for p=1,q=0.

3. Originally Posted by kompik
I think that your solution is ok. You may notice that the last line is in fact $\displaystyle p\Rightarrow q$.

You can also use truth table to verify that your answer is correct.
For proposition which have the form of implication there is a shorter way:
Clearly $\displaystyle p\Rightarrow q$ if false only in one case: p=1, q=0.
Let us try to the same thing with the first expression
$\displaystyle [p\land \neg(p\Rightarrow q)] \Rightarrow q$.
This implication is false only if [tex][p\land \neg(p\Rightarrow q)]=1[/math and q=0.
Kompik,