[SOLVED] Boolean Simplification

Printable View

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 $p\Rightarrow q$ .

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 $p\Rightarrow q$ if false only in one case: p=1, q=0.
Let us try to the same thing with the first expression
$[p\land \neg(p\Rightarrow q)] \Rightarrow q$ .
This implication is false only if $[p\land \neg(p\Rightarrow q)]=1$ and q=0. For a conjunction to be true, you need that both of the operands are true. This means that p=1, and $\neg(p\Rightarrow q)$ 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 $p\Rightarrow q$ .

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 $p\Rightarrow q$ if false only in one case: p=1, q=0.
Let us try to the same thing with the first expression
$[p\land \neg(p\Rightarrow q)] \Rightarrow q$ .
This implication is false only if [tex][p\land \neg(p\Rightarrow q)]=1[/math and q=0.

Kompik,

Thanks for your reply. Much appreciated. :)