# Showing a conclusion is correct (with deduction/reduction)

• Sep 27th 2009, 12:22 PM
JacobSkylar
Showing a conclusion is correct (with deduction/reduction)
I'll use some replacement characters since I don't know how to make the correct mathematical ones.

A = Logic symbol for and
--> = symbol for implies
- (it's that hook thing normally) =
negative

Here's the problem:

Show that the conclusion

(p $\rightarrow$ q) $\wedge$ (-p $\rightarrow$ r) $\wedge$ ((-p $\wedge$ r) $\rightarrow$ s) $\wedge$ -q ==> s

is correct. Do this using both deduction and reduction methods.

I'm just stranded here, I really don't know where to begin and how to proceed. Thanks for any help.
• Sep 27th 2009, 12:28 PM
Plato
Quote:

Originally Posted by JacobSkylar
I'll use some replacement characters since I don't know how to make the correct mathematical ones.

A = Logic symbol for and
--> = symbol for implies
- (it's that hook thing normally) =
negative

Here's the problem:

Show that the conclusion

(p --> q) A (-p --> r) A ((-p A r) --> s) A -q ==> s

is correct. Do this using both deduction and reduction methods.

I'm just stranded here, I really don't know where to begin and how to proceed. Thanks for any help.

You can learn to use symbols
$$\wedge$$ gives $\wedge$ is and
$$\Rightarrow$$ gives $\Rightarrow$ is imply
• Sep 27th 2009, 12:36 PM
JacobSkylar
Quote:

Originally Posted by Plato
You can learn to use symbols
$$\wedge$$ gives $\wedge$ is and
$$\Rightarrow$$ gives $\Rightarrow$ is imply

Fixed.
• Sep 27th 2009, 12:48 PM
Plato
Quote:

Originally Posted by JacobSkylar
(p $\rightarrow$ q) $\wedge$ (-p $\rightarrow$ r) $\wedge$ ((-p $\wedge$ r) $\rightarrow$ s) $\wedge$ -q ==> s

$\neg q$ from the first gives $\neg p$.
$\neg p$ gives $r$ which gives $\neg p\wedge r$.
That in turn gives $s$
• Sep 27th 2009, 12:57 PM
JacobSkylar
Quote:

Originally Posted by Plato
$\neg q$ from the first gives $\neg p$.
$\neg p$ gives $r$ which gives $\neg p\wedge r$.
That in turn gives $s$

Which of the two methods was that, and would you mind showing the other one too? Feel free to add in why things do what they do, what rules you apply.

Thanks alot.
• Sep 27th 2009, 01:08 PM
Plato
Quote:

Originally Posted by JacobSkylar
Which of the two methods was that, and would you mind showing the other one too? Feel free to add in why things do what they do, what rules you apply.

I don't know what you mean.
I used simplification to get $\neg q$.
Then modus tollens to get $\neg p$.
Then modus ponens to get $r$.
Then modus conjunction to get $\neg p\wedge r$.
Then modus ponens to get $s$.
• Sep 27th 2009, 01:10 PM
JacobSkylar
Quote:

Originally Posted by Plato
I don't know what you mean.
I used simplification to get $\neg q$.
Then modus tollens to get $\neg p$.
Then modus ponens to get $r$.
Then modus conjunction to get $\neg p\wedge r$.
Then modus ponens to get $s$.

Okay, thanks alot.

Nvm, I realised you used the deduction method. Would you know how to solve it using the reduction method too?
• Sep 30th 2009, 12:37 PM
JacobSkylar
So, do you know how to solve it using the reduction method?

1. Assume that the conclusion is false
2. The premises are true but the conclusion is false
3. This would mean http://www.mathhelpforum.com/math-he...b892adae-1.gif is true etc etc

(note I don't know the mathematical terms in english for everything)

Would be really appreciated.