YAM(ath)T: propositional logic, I absolutely don't get it

[ndee]

Here's the thing:

(p --> q) /\ (q --> r) --> (p --> r)

the /\ stands for the AND and the --> for implication. Our prof gave us this assingment and I need to make a truth table for that. I did read the theory about it but I don't get it. Can I just put some random values into p, q and r?

 

[tweakmm]

You don't understand how p-->r or you dont know how to make a truth table?

 

[GreatBarracuda]

Originally posted by: ndee
Here's the thing:

(p --> q) /\ (q --> r) --> (p --> r)

the /\ stands for the AND and the --> for implication. Our prof gave us this assingment and I need to make a truth table for that. I did read the theory about it but I don't get it. Can I just put some random values into p, q and r?

easy....use the follwing tables:

a b a->b
--------------
0 0 1
0 1 1
1 0 0
1 1 1


a b a^b
0 0 0
0 1 0
1 0 0
1 1 1

 

[ndee]

tweakmm:
I know how to make a truth table, for every possible combination the "result" but I don't know why or how I can get to p --> r.

 

[TuxDave]

I figure you make a list of all the combinations of p,q,r... find the truth table for p-->q, q-->r,p-->r and find if all the cases where the first two are both true, is the last one true?

 

[Howard]

too lazy to explain properly

 

[ndee]

Originally posted by: TuxDave
I figure you make a list of all the combinations of p,q,r... find the truth table for p-->q, q-->r,p-->r and find if all the cases where the first two are both true, is the last one true?

I just think it really depends on what values you choose for p, q and r.

GreatBarracuda:
how come when a = 1 and b = 0, that a->b is 0?

 

[GreatBarracuda]

Originally posted by: ndee
how come when a = 1 and b = 0, that a->b is 0?

That's how it is.

Check this

 

[eigen]

Are you asking more of how to derive the final clause from hypothesis?

 

[clamum]

Here's the truth table for the implication operator. Assume p --> q

If p is false and q is false, the whole thing is true.
If p is false and q is true, the whole thing is true.
If p is true and q is false, the whole thing is false.
If p is true and q is true, the whole thing is true.

To make the truth table for your problem, make a column each for p, q, and r. Put all possible values for the three vars in there, you should have 2^3 = 8 rows.
Then make a column for p --> q and put the result of the implication for each row.
Then make a column for q --> r " "
Then make a column for p --> r " "
Then make a column for (p --> q) /\ (q --> r) and put the result of the AND for each row.
Finally, make one for (p --> q) /\ (q --> r) --> (p --> r) and put the result of the implication for each row.

 

[ndee]

Originally posted by: clamum
Here's the truth table for the implication operator. Assume p --> q

If p is false and q is false, the whole thing is true.
If p is false and q is true, the whole thing is true.
If p is true and q is false, the whole thing is false.
If p is true and q is true, the whole thing is true.

To make the truth table for your problem, make a column each for p, q, and r. Put all possible values for the three vars in there, you should have 2^3 = 8 rows.
Then make a column for p --> q and put the result of the implication for each row.
Then make a column for q --> r " "
Then make a column for p --> r " "
Then make a column for (p --> q) /\ (q --> r) and put the result of the AND for each row.
Finally, make one for (p --> q) /\ (q --> r) --> (p --> r) and put the result of the implication for each row.

The Problem with grasping is that I can think of a practical problem but when I only get p and q, I can't imagine something under those 2 variables.

 

[Kibbo]

For p-->q, where p=1 and q=0, here's an example.

If the sky is blue, then we are on mars.

The sky is blue (p=1) but we are not on mars (q=0). Thus, the truth value of the statement is 0, because the whole implication is false (obviously).