Simple Logic Truth Table Needs Checking

[Hotsuma]

Dear All,

Having trouble with a seemingly simple logic truth table. Are these answers correct?

$$\begin{tabular}{| c | c | c | c | c | c |}<br /> \hline<br /> p & q & r & (p \vee q)\wedge(q \vee r) & (\neg p \wedge q) \vee ( p \wedge \neg r) & p \rightarrow q \rightarrow r\\<br /> \hline<br /> T & T & T & T & F & T \\<br /> T & T & F & T & T & F \\<br /> T & F & T & T & F & T \\<br /> T & F & F & F & T & F \\<br /> F & T & T & T & T & T \\<br /> F & T & F & T & T & F \\<br /> F & F & T & F & F & T \\<br /> F & F & F & F & F & T \\<br /> \hline<br /> \end{tabular}$$

 

[lanedance]

Hi Hotsuma, I think the and/or/not parts are ok, but for the last part I think it would be

$$\begin{tabular}{| c | c | c | c } <br /> \hline <br /> p & q & r & p \rightarrow q \rightarrow r\\<br /> \hline<br /> T & T & T & T \\<br /> T & T & F & F \\<br /> T & F & T & F \\<br /> T & F & F & F \\<br /> F & T & T & T \\<br /> F & T & F & T \\<br /> F & F & T & T \\<br /> F & F & F & T \\<br /> \hline\end{tabular}$$

This effectively has 2 tests, it first requires that if p then q, if this is satisfied the the further condition that if q then r must be satisfied

In the 3rd case, as p is T and q is F, then the line is FALSE as the first condition is not satisfied

For the last 4 cases as p is F then the line is T by default, and we don't go any further - its a vacuous truth

what do you think?

 

[Hotsuma]

Dear lanedance,

I considered this but wasn't sure if that was correct or not. What you described to me in your past post could be written as "if p then q" and "if q then r," which I don't think is the same thing, or rather, I'm not sure whether it is or is not. Thanks for the suggestion though, having someone else recommend what I was thinking is always appreciated.

 

[Hotsuma]

And I'd love to prove that the two are tautologies, but since I am not sure about the former, that doesn't really help me much :(.


$$(p \rightarrow q) \wedge (q \rightarrow r) \equiv p \rightarrow q \rightarrow r$$

 

[lanedance]

Dear lanedance,

I considered this but wasn't sure if that was correct or not. What you described to me in your past post could be written as "if p then q" and "if q then r," which I don't think is the same thing, or rather, I'm not sure whether it is or is not.

I don't think is is exactly the same thing, take the case:
p = F
q = T
r = F

this gives the following:

$$p \rightarrow q$$
True - vacuously as P is False

$$q \rightarrow r$$
False

$$(p \rightarrow q) \wedge (q \rightarrow r)$$
False

$$p \rightarrow q \rightarrow r$$
True - vacuously as P is False

 

[Elucidus]

There is a problem here. Implication is not an associative connective. i.e.

$$(p \rightarrow q) \rightarrow r \not\equiv p \rightarrow (q \rightarrow r)$$

so the meaning of $p \rightarrow q \rightarrow r$ is ambiguous.

(Consider the case where p = F, q = T, r = F.)

Unless there is an assumption about the order of execution of multiple implications, the last column could be evaluated differently (either TFTTTFTF or TFTTTTTT).

--Elucidus

 

[Hotsuma]

Yeah, that was my fear. I e-mailed the professor at like 4PM but he has not yet responded (nor do I expect him to).