# 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.