Truth table help... did I do it correctly

1. Jan 28, 2016

Kingyou123

1. The problem statement, all variables and given/known data

2. Relevant equations
N/A

3. The attempt at a solution
My proof table, I'm not sure but it seems that PΞQ is not true.

2. Jan 28, 2016

Buzz Bloom

Hi Kingyou:

Your calculation looks right to me.

Regards,
Buzz

3. Jan 28, 2016

MrAnchovy

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
Code (Text):
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
Code (Text):
a b | a→b
----------
T T | T
T F | F
F T | T
F F | T

4. Jan 28, 2016

Kingyou123

Awesome, thank you

5. Jan 29, 2016

WWGD

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.

6. Jan 29, 2016

MrAnchovy

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

7. Jan 29, 2016

WWGD

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

8. Jan 29, 2016

MrAnchovy

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?

9. Jan 29, 2016

WWGD

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

10. Jan 29, 2016

MrAnchovy

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

11. Jan 29, 2016

WWGD

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 $p \rightarrow - r$..

Last edited: Jan 29, 2016
12. Jan 30, 2016

MrAnchovy

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

13. Jan 30, 2016

WWGD

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.

14. Jan 30, 2016

MrAnchovy

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

15. Feb 2, 2016

Kingyou123

What is wrong with my work?

16. Feb 2, 2016

MrAnchovy

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.

17. Feb 2, 2016

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 $(p \to r)$.

18. Feb 3, 2016

SammyS

Staff Emeritus
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.