# [SOLVED] Boolean Simplification

• May 2nd 2010, 03:59 AM
Ashgasm
[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!
• May 2nd 2010, 04:14 AM
kompik
Quote:

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.
• May 2nd 2010, 04:19 AM
Ashgasm
Quote:

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,