Help! Conditional proof question

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Reply Sun 19 Jul, 2009 03:08 pm
I was rolling along just fine in my logic class until we got to conditional proofs. I understand the rules on their own. I understand where I'm starting and where I need to go; but when I look at the premises, for some reason I cannot see the path to get me to my conclusion. I will be posting homework help in the homework forum. I would appreciate any help.

I guess I can't post anywhere else yet. Can someone please help me.

1. ~KvL Premise
2. K->(L->M) /KvM Premise

When I look at this, I immediately think that I should assume K. Which then gives me (L->M) by MP. Then if I nest the other conditional, I would assume L which gives me M by MP. But then I have K->M which is not what I need when I use Imp because it gives me ~KvM. Help!
 
kennethamy
 
Reply Mon 20 Jul, 2009 06:38 am
@oldmanhood,
oldmanhood;78337 wrote:
I was rolling along just fine in my logic class until we got to conditional proofs. I understand the rules on their own. I understand where I'm starting and where I need to go; but when I look at the premises, for some reason I cannot see the path to get me to my conclusion. I will be posting homework help in the homework forum. I would appreciate any help.

I guess I can't post anywhere else yet. Can someone please help me.

1. ~KvL Premise
2. K->(L->M) /KvM Premise

When I look at this, I immediately think that I should assume K. Which then gives me (L->M) by MP. Then if I nest the other conditional, I would assume L which gives me M by MP. But then I have K->M which is not what I need when I use Imp because it gives me ~KvM. Help!


~KvM
K (Assumption)
M (DS)
MvK (Addition)
KvM (Commutation)
 
mickalos
 
Reply Mon 20 Jul, 2009 07:05 am
@oldmanhood,
oldmanhood;78337 wrote:
I was rolling along just fine in my logic class until we got to conditional proofs. I understand the rules on their own. I understand where I'm starting and where I need to go; but when I look at the premises, for some reason I cannot see the path to get me to my conclusion. I will be posting homework help in the homework forum. I would appreciate any help.

I guess I can't post anywhere else yet. Can someone please help me.

1. ~KvL Premise
2. K->(L->M) /KvM Premise

When I look at this, I immediately think that I should assume K. Which then gives me (L->M) by MP. Then if I nest the other conditional, I would assume L which gives me M by MP. But then I have K->M which is not what I need when I use Imp because it gives me ~KvM. Help!


It's not clear from this what it is you are trying to prove. Is it supposed to be ~KvL, K->(L->M) |- KvM ? This isn't a valid argument. Counterexample:
K = F
L = F
M = F
 
kennethamy
 
Reply Mon 20 Jul, 2009 09:39 am
@mickalos,
mickalos;78400 wrote:
It's not clear from this what it is you are trying to prove. Is it supposed to be ~KvL, K->(L->M) |- KvM ? This isn't a valid argument. Counterexample:
K = F
L = F
M = F


The values you assign give us true premises and a true conclusion. So why isn't the argument valid?
 
mickalos
 
Reply Mon 20 Jul, 2009 10:52 am
@kennethamy,
kennethamy;78426 wrote:
The values you assign give us true premises and a true conclusion. So why isn't the argument valid?


If the argument you are trying to prove is indeed:

~KvL, K->(L->M) |- KvM

And as you didn't state otherwise, I assume it is, then my counterexample is correct. (KvM) is false if and only if both K and M are false, the truth table for the disjunction looks like this:
http://onegoodmove.org/fallacy/images/or.gif
 
kennethamy
 
Reply Mon 20 Jul, 2009 12:23 pm
@mickalos,
mickalos;78434 wrote:
If the argument you are trying to prove is indeed:

~KvL, K->(L->M) |- KvM

And as you didn't state otherwise, I assume it is, then my counterexample is correct. (KvM) is false if and only if both K and M are false, the truth table for the disjunction looks like this:
http://onegoodmove.org/fallacy/images/or.gif

So if K is false, then ~K is true, and ~KvM, is true.
 
mickalos
 
Reply Mon 20 Jul, 2009 12:55 pm
@kennethamy,
kennethamy;78446 wrote:
So if K is false, then ~K is true, and ~KvM, is true.


Yes, and an argument is invalid if the premises can all be true while the conclusion is false. KvM is your conclusion isn't it?

Whichever way you arrange that set of sentences, you won't reach a valid argument.

KvM, K->(L->M) |- ~KvL Counterexample:

K= T
L = F
M = T

KvM, ~KvL |- K->(L->M) Counterexample:

K = T
M = F
L = T
 
patriciabeth
 
Reply Thu 13 Aug, 2009 02:10 am
@oldmanhood,
(H & S) ⊃ ~ (F ≡ S)

I too am having problems with determining the proof for this.......please help!
patriciabeth
 
mickalos
 
Reply Thu 13 Aug, 2009 08:26 am
@patriciabeth,
patriciabeth;82913 wrote:
(H & S) ⊃ ~ (F ≡ S)

I too am having problems with determining the proof for this.......please help!
patriciabeth


You can't prove that sentence without having some premises (e.g (H & S) -> ~F), as it isn't a tautology. The sentence is false when:
H = T
S = T
F = T
 
 

 
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