# Math Help - Logical statements

1. ## Logical statements

Peter states the following facts about the football premiership:
If St Kilda beat Hawthorn, Franklin is included next week
Hawthorn beating St Kilda means that Geelong will top the ladder
Hawthorn will make the top eight only if Carlton miss out on the finals
If Geelong lose to Essendon then Geelong can't top the ladder
If Hawthorn fail to make the top eight, Franklin will be dropped next week

a) Formulate these five facts as statements in propositional logic.

I did this:

p: St Kilda beat Hawthorn
q: Franklin included next week
y: Hawthorn will make the top 8
z: Carlton miss out on finals
a: Geelong lose to Essendon

1. $p \to q$
2. $\neg p \to x$
3. $y \leftrightarrow z$
4. $a \to \neg x$
5. $\neg y \to \neg q$

b) Is it the case that if Geelong don't manage to top the ladder, Carlton miss out
on the finals? Give your reasoning.

So is this part asking whether $\neg x \to z$ is true or false?

Do I use rules of inference from my 5 propositions from part a) to figure out b)? I can't seem to link them together...

Any help would be much appreciated!

2. Deleted

3. ?

4. I posted something, but I decided it had been too long since I have done this and wasn't confident in the solution so I deleted it.

5. Ohhh right, haha, sorry, I thought you meant I had to delete my post lolz

6. Ok I 'think' I got it after many trials, can anyone confirm if my working is correct/wrong. Thanks very much!

From 1. $p \to q \equiv \neg q \to \neg p$ name this 1'.
Assuming 5 and 1' are true and then using Hypothetical Syllogism we yield $\neg y \to \neg p \equiv p \to y$ to be true.

From 3. $y \leftrightarrow z \equiv (y \to z) \land (z \to y)$ name this 3'. Assuming 3' is true then using Simplification yields $y \to z$ to be true.

Looking at $p \to y$ and $y \to z$ using Hypothetical Syllogism we yield $p \to z$ to be true.

From 2. we have $\neg p \to x \equiv \neg x \to p$ name this 2'.

Using Hypothetical Syllogism with 2' and $p \to z$ yields $\neg x \to z$.

Thus $\neg x \to z$ is true!

7. I guess I could have left my answer, but yours is prettier.

8. Referring to the OP, your statement 3 is incorrect. It should be

$z\to y.$

The "only if" keyword works like this: A only if B gets translated $B\to A.$ The bidirectional arrow is the translation of "if and only if".

9. Originally Posted by Ackbeet
The "only if" keyword works like this: A only if B gets translated $B\to A.$

The bit above in red is incorrect.
It should be: $\text{“A only if B”} \equiv \left( {A \to B} \right)$.

Reason: It says that A is true only if B is also true.
The only condition we cannot allow is true implies false.

10. Thanks for the help guys!

Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?

edit - ah just as I posted this, plato confirmed it haha

11. Originally Posted by usagi_killer
Thanks for the help guys!

Ackbeet, thanks for that pickup, that was a stupid mistake by me - however I was under the impression that "A if B" means B implies A, however "A only if B" means A implies B?

edit - ah just as I posted this, plato confirmed it haha
Yes, because that is where if and only if comes from, i.e., $A\to B \wedge B \to A \equiv A \leftrightarrow B$

12. Hmmm, it seems I am stuck with a further extension of this problem.

The extension is that "can one deduce that either Geelong lose to Essendon or Carlton make the finals?"

So my interpretation of this question is that is $a \lor \neg z$ true or false?

Now to prove this, if $a$ is true or $\neg z$ is true then $a \lor \neg z$ is true, however if $a$ and $\neg z$ are both false, then $a \lor \neg z$ is false.

But no matter how hard I try, I can not seem to find any way of manipulating the information we have to prove what I described above

13. Reply to Plato: Yeah, I was thinking of "A if B". Sorry about that.

14. Originally Posted by usagi_killer
Hmmm, it seems I am stuck with a further extension of this problem.

The extension is that "can one deduce that either Geelong lose to Essendon or Carlton make the finals?"

So my interpretation of this question is that is $a \lor \neg z$ true or false?

Now to prove this, if $a$ is true or $\neg z$ is true then $a \lor \neg z$ is true, however if $a$ and $\neg z$ are both false, then $a \lor \neg z$ is false.

But no matter how hard I try, I can not seem to find any way of manipulating the information we have to prove what I described above
I think that inference is invalid.

When P: T, Q: T, X: T, Y: T, Z: T, A: F, then the argument is such that all the premises are true, and the conclusion is false. Therefore, the argument is invalid. Use a truth table to check.

15. Thanks for your help but I don't quite understand how to solve it yet.

I did this:

$a \lor \neg z \equiv z \to a$

So now I have changed it to $z \to a$ I don't know how to show whether that is true or false.

Page 1 of 2 12 Last