SD Theorems Logic HW due Please help

  • Thread starter yankes2k
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In summary, Tony needs help proving a theorem for logic and is lost. He asks for help and thanks those who offer help. Tony has rules for conditional introduction, disjunction introduction, conjunction elimination, negation elimination/introduction, and conditional elimination, but does not know how to use the Commutation and Exportation rules. He types out the rules, but is still lost. He eventually proves the rules himself.
  • #1
yankes2k
11
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Hi everyone,

I really need help proving a theorem for logic HW. I am allowed to use all the standard derivations in Sentential Logic.

Also I do not know how to enter in the symbols so I will use ">" to signify conditional in the problem below.

Instructions: Show that each of the following is a theorem in SD by constructing a derivation.

1:) ~A>((B&A)>C)

2:) (AvB)>(BvA)

If someone could help me out I would really appreciate it as I am so lost right now.

Thanks,

-Tony
 
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  • #2
1:) ~A>((B&A)>C)
Assume ~A. Use Commutation and Exportation ([(p & q) -> r] <=> [p -> (q -> r)]) on [(B & A) -> C] to get a new statement. See it now?

2:) (AvB)>(BvA)
Eh, this is just Commutation. Do you not have a rule that says [(p v q) <=> (q v p)]?
Edit: Okay, that's probably a stupid question on my part. Can you type up the rules you have? I'm guessing you have Modus Ponens, Modus Tollens,
(p & q) => p
p, q, => (p & q)
p => p v q
What else?
 
Last edited:
  • #3
I can use conjunction elimination and introduction
biconditional introduction and elimination
negation elimination/introduction
disjunction elimination/introduction
conditional introduction and elimination

that's it.
 
  • #4
Also I don't know how to use "Commutation and Exportation" rules either. We can only use what is listed above. Class is at 1:30 U.S. Eastern standard time please help!
 
  • #5
yankes2k said:
I can use conjunction elimination and introduction
biconditional introduction and elimination
negation elimination/introduction
disjunction elimination/introduction
conditional introduction and elimination

that's it.
I have to go look up what these rules are- they go by different names. If you really want to save time, you could type out the rules.
 
  • #6
conjunction ibtroduction
P
Q
P&Q

conjunction Elination
P&Q
P or Q

biconditional introduction

Sub Der:Assume P
Q

Assume Q
P
P (triple Bar) Q

biconditional Elimination
P (triple bar) Q
P or Q
THerefore, Q Therefore, P

Disjunction Intro

P

PvQ or QvP

Disjunction Elination

PvQ
sub Derivation: P
R

Sub Derivation:Q
R
Therefore, R

Negation introduction

Subderivation:P
Q
~Q
Therefore, ~P

Negation Elimination
Sub derivation:~P
Q
~Q
THerefore, P
 
  • #7
Sorry forgot conditional intro and elimination

Condition elimination

P implies Q
P
Therefore, Q

Condition introduction
Subderivation:P
Q
Therefore, P implies Q
 
  • #8
Okay, I'm used to having more rules- every proof I think of uses rules you don't have. While I'm thinking, here's what I meant earlier:
1) (B & A) -> C
2) (A & B) -> C [1, Commutation]
3) A -> (B -> C) [2, Exportation]
4) ~A v (B -> C) [3, Implication]

So you would just go backwards. In your proof, you assume ~A (hypothetically) and try to derive (B & A) -> C, so you can infer ~A -> (B & A) -> C.

1) ~A [conditional introduction]
2) ~A v (B -> C) [1, disjunction introduction]
3)

Maybe you can think of a way to do this using your rules.
 
  • #9
I'll just name the rules that you need, and maybe you can prove them yourself. Or maybe you have already proven them.

(P -> Q) <=> (~P v Q)
(P -> Q) <=> (~Q -> ~P)
(~~P) <=> P

If you can prove those:

1)) A v B [conditional intro]
2)) ~~A v B [1, subproof (~~P) <=> P]
3)) ~A -> B [2, subproof (P -> Q) <=> (~P v Q)]
4)) ~B -> ~~A [3, subproof (P -> Q) <=> (~Q -> ~P)]
5)) ~B -> A [4, subproof (~~P) <=> P]
6)) B v A [5, subproof (P -> Q) <=> (~P v Q)]
7) (A v B) -> (B v A) [1, 6, conditional intro]
 
  • #10
Are you still there?

(P & Q) <=> (Q & P) is easy as pie
1)) P & Q
2)) Q
3)) P
4)) Q & P
5) (P & Q) -> (Q & P)
Repeat other way. So you have commutation for conjunction: (P & Q) <=> (Q & P)

Exportation the way you need it is also easy as pie, if you have (P -> Q) <=> (~P v Q):

1)) P -> (Q -> R) [1, conditional intro)
2)) ~P v (Q -> R) [2, subproof (P -> Q) <=> (~P v Q)]
3)) ~P v (~Q v R) [3, subproof (P -> Q) <=> (~P v Q)]
4)) (~P v ~Q) v R [4, Please say you've proved association?]
5)) ~(P & Q) v R [5, and DeMorgan's Theorems?]
6)) (P & Q) -> R [6, subproof (P -> Q) <=> (~P v Q)]
7) (P -> (Q -> R)) -> ((P & Q) -> R [1, 6]

Have you seen these before? Am I just scaring you?
 

1. What is SD Theorems Logic?

SD Theorems Logic is a branch of mathematical logic that deals with the study of formal systems and their properties. It is used to prove the validity of logical arguments and to analyze the structure of mathematical proofs.

2. What are some common applications of SD Theorems Logic?

SD Theorems Logic is used in various fields such as computer science, philosophy, linguistics, and artificial intelligence. It is also commonly used in the development and analysis of programming languages and software.

3. What does HW due mean in the context of SD Theorems Logic?

HW due stands for homework due, and it refers to any assignments or tasks related to SD Theorems Logic that are due on a specific date. These can include problem sets, exercises, or projects assigned by a teacher or professor.

4. How can I get help with my SD Theorems Logic homework?

If you need help with your SD Theorems Logic homework, the best place to start is by consulting your textbook, class notes, or online resources. You can also reach out to your teacher or classmates for clarification or additional assistance.

5. Is SD Theorems Logic difficult to learn?

SD Theorems Logic can be challenging, but it is a fundamental subject in mathematics and computer science. With dedication and practice, anyone can learn SD Theorems Logic and apply it to various fields and real-world problems.

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