Truth table help.... did I do it correctly

[Kingyou123]

Homework Statement

The question I'm asking about is number two. Should've made it clearer.

Homework Equations

N/A

The Attempt at a Solution

My proof table, I'm not sure but it seems that PΞQ is not true.

 

[Buzz Bloom]

Hi Kingyou:

Your calculation looks right to me.

Regards,
Buzz

 

[pbuk]

A few hints:

1. Set out your truth table methodically - this makes it easier for you to check and easier for a marker to give you partial marks
   

   p q r
   -----
   T T T
   T T F
   T F T
   T F F
   F T T
   F T F
   F F T
   F F F

2. Read the question carefully - why have you calculated p → r ?
3. Learn the truth table for implies
   

   a b | a→b
   ----------
   T T | T 
   T F | F
   F T | T
   F F | T

 

[Kingyou123]

Hi Kingyou:

Your calculation looks right to me.

Regards,
Buzz

Awesome, thank you

 

[WWGD]

Homework Statement

The question I'm asking about is number two. Should've made it clearer.
View attachment 94964

Homework Equations

N/A

The Attempt at a Solution

My proof table, I'm not sure but it seems that PΞQ is not true.
View attachment 94963

You can also use an argument:
Assume $p\rightarrow$ ~ r , and assume p. Then you have $p\rightarrow q$, from which q follows ,from which r follows.

 

[pbuk]

You can also use an argument:
Assume $p\rightarrow$ ~ r , and assume p. Then you have $p\rightarrow q$, from which q follows ,from which r follows.

How does that show that there are values of (p, q, r) for which the identity does not hold)?

 

[WWGD]

How does that show that there are values of (p, q, r) for which the identity does not hold)?

It shows there are none, since this shows that the wff is a theorem.

 

[pbuk]

It shows there are none, since this shows that the wff is a theorem.

I don't understand which wff you are referring to, or any steps of your outline proof I am afraid. In any case I'm pretty sure the question setter is looking for a truth table and is going to mark a deductive proof harshly unless it is presented flawlessly - why take the risk?

 

[WWGD]

I don't understand which wff you are referring to, or any steps of your outline proof I am afraid. In any case I'm pretty sure the question setter is looking for a truth table and is going to mark a deductive proof harshly unless it is presented flawlessly - why take the risk?

Sorry, I was unclear, I was aiming for a proof by contradiction, but I agree, might as well go for the truth table argument.

 

[pbuk]

1. 
   
2. Read the question carefully - why have you calculated p → r ?

To be clear, I meant why have you calculated p → r (incorrectly) in the first calculated column of the TT?

 

[WWGD]

To be clear, I meant why have you calculated p → r (incorrectly) in the first calculated column of the TT?

I assumed $p \rightarrow -r$ together with the given premises and concluded r , so I showed by contradiction that $p \rightarrow r$. Unfortunately I don't know well the Latex for logic symbols to do
the actual derivation of $p \rightarrow r$ and $p \rightarrow - r$..

 

[pbuk]

Oh this has got very confusing, my post #10 was intended to clarify my post #3 pointing out that the OP had misread the question which asks for (p → q) → (q → r) ≡ (p → r) and attempted to show a TT for (p → r) → (q → r) ≡ (p → r) instead.

As for your post #11 WWGD, note that the negation of $(p \to q) \to (q \to r) \equiv (p \to r)$ is not $(p \to q) \to (q \to r) \equiv (p \to \neg r)$ it is $(p \to q) \to (q \to r) \neq (p \to r)$

 

[WWGD]

Oh this has got very confusing, my post #10 was intended to clarify my post #3 pointing out that the OP had misread the question which asks for (p → q) → (q → r) ≡ (p → r) and attempted to show a TT for (p → r) → (q → r) ≡ (p → r) instead.

As for your post #11 WWGD, note that the negation of $(p \to q) \to (q \to r) \equiv (p \to r)$ is not $(p \to q) \to (q \to r) \equiv (p \to \neg r)$ it is $(p \to q) \to (q \to r) \neq (p \to r)$

I know (and you're right that I only proved one side of the equivalence), I don't mean to be impolite, but it is getting too confusing; let's just drop it if you don't mind, sorry for the dead end.

 

[pbuk]

it is getting too confusing; let's just drop it if you don't mind

Agreed - the OP seems to have gone anyway (unfortunately I think with the impression that his workings were OK but we tried...)

 

[Kingyou123]

Agreed - the OP seems to have gone anyway (unfortunately I think with the impression that his workings were OK but we tried...)

What is wrong with my work?

 

[pbuk]

Look at what you have written at the top of your fourth and sixth columns and read the question again.

And then apply the truth table I showed you correctly.

 

[mfig]

The easiest way to do this is to create the truth table for the expression

$(p \to q) \to (q \to r)$

Then look for two things. First, check to see if the truth value (for any fixed values of p and r) depends on q. If so, the equivalence cannot hold. If not, then check if the truth table for the above expression produces the same values for a fixed p and r as $(p \to r)$.

 

[SammyS]

What is wrong with my work?

You have the wrong result for a ⇒ b . The only instance in which a ⇒ b is false, is the case of a is false and b is true.