# Logic Proofs

• Feb 10th 2011, 01:50 PM
Newskin01
Logic Proofs
Hello,

I'm not getting credit for this questions, but was told there would be one like it on the quiz so I'm trying to figure it out.

Write a proof sequence for the following assertion.

http://www.mathhelpforum.com/math-he...oofs-proof.jpg

I know I am given several things I just not sure how to start a proof. We kind of just were thrown in this and never having experience with it is making it a confusing but like any math I know practice will make me better, but first I need to understand how this work. Not only are proofs something new I have to do but the logic is all new so that doesn't help. I'd appreciate any help that can be given. I know it's not your job to do my work so hints are acceptable help.

Thanks.
• Feb 10th 2011, 02:14 PM
Plato
Those are unknown logical symbols, at least to me.
Please tell us what each means.
• Feb 10th 2011, 02:19 PM
Newskin01
¬ = not
--> is if then
^ = and
=> is equivalent

I hope this helps. Thanks for looking.
• Feb 10th 2011, 02:25 PM
Plato
Quote:

Originally Posted by Newskin01
¬ = not
--> is if then
^ = and
=> is equivalent

Those symbols are completely standard.
BUT they are not in your original posting.
Please edit the post to rid it of any special fonts which do not display very well.
• Feb 10th 2011, 03:29 PM
Soroban
Hello, Newskin01!

Did you even look at what you posted?

Quote:

Write a proof sequence for the following assertion.

. . $\displaystyle \bigg[\sim(\sim\!p \to q) \vee (\sim\!p \:\wedge \sim\!q)\bigg] \:\Longleftrightarrow\:(\sim\!p \:\wedge \sim\!q)$

Is this what you meant?

It is a rather silly statement, isn't it?

The first part is: .$\displaystyle \sim(\sim\!p \to q) \;\;=\;\;\sim(p \vee q) \;\;=\;\;\sim\!p\:\wedge \sim\!q$

The statement becomes:
. . $\displaystyle \bigg[(\sim\!p \:\wedge \sim\!q) \vee (\sim\!p \,\wedge \sim\!q)\bigg] \;\Longleftrightarrow\; (\sim\!p \:\wedge \sim\!q)$

which simplifies to:
. . . . . . . . . . . . . $\displaystyle (\sim\!p \:\wedge \sim\!q) \;\Longleftrightarrow\;(\sim\!p \:\wedge \sim\!q)$

which is true, of course.

• Feb 10th 2011, 03:35 PM
Plato
Quote:

Originally Posted by Newskin01
Hello,

I'm not getting credit for this questions, but was told there would be one like it on the quiz so I'm trying to figure it out.

Write a proof sequence for the following assertion.

¬(¬p-->q) "or" (¬p^¬q) => ¬p^¬q

Is it just my browser, can anyone else actually read this positing?
• Feb 10th 2011, 03:38 PM
Newskin01
Oh it doesn't look that way on my screen weird. I can post a picture to help. Attachment 20742
• Feb 10th 2011, 03:54 PM
Plato
Quote:

Originally Posted by Newskin01
Oh it doesn't look that way on my screen weird. I can post a picture to help. Attachment 20742

When posting do not use special fonts.
Not all browsers are equal.
• Feb 10th 2011, 04:08 PM
Newskin01
I'm sorry. Will not happen again.
• Feb 10th 2011, 06:50 PM
Ackbeet
I'm going to trot out emakarov's standard question here: what rules of inference are you using? Natural deduction/Fitch style? Copi's 19 Rules? Something else?
• Feb 10th 2011, 06:59 PM
Newskin01
Quote:

I'm going to trot out emakarov's standard question here: what rules of inference are you using? Natural deduction/Fitch style? Copi's 19 Rules? Something else?
Natural deduction, didn't realize there were so many. I'm just starting in on the topic.

Thanks
• Feb 11th 2011, 01:42 AM
emakarov
Here is a derivation of $\displaystyle \neg p\land\neg q$ from an open assumption $\displaystyle \neg(\neg p\to q)$ in natural deduction. From here, it is easy to construct the required derivation using the disjunction elimination rule.

I assume that $\displaystyle \bot$ is a primitive symbol (for falsehood) and $\displaystyle \neg p$ is a contraction for $\displaystyle p\to\bot$. If you have other conventions and rules for negation, feel free to describe them. The labels in implication introduction rules show the assumptions that are closed by these rules.