0
$\begingroup$

I'm given a conditional statement

~p ∨ q -> r

From what I understand, the state of r is dependent on the state of p and q. The truth table I came up with based on this is: my truth table EDIT: the r on the bottom row should be 1. That was a typo.

However, I used a truth table generator to check my answer, and I'm getting confusing results.

Confusing Results

I fail to understand how r can have different states despite p and q remaining the same, if its conditionally dependent on them. I also don't understand where the constants on the rightmost column are being derived from. Wouldn't they be the same as r?

Am I just using the online truth table tools wrong?

$\endgroup$
  • $\begingroup$ And what is the question you've been asked? $\endgroup$ – amrsa Feb 13 '18 at 20:52
  • $\begingroup$ @amrsa To write a truth table for the conditional statement "~p ∨ q -> r" $\endgroup$ – user2651000 Feb 13 '18 at 20:55
  • $\begingroup$ That's not what you did. The answer to that is the other link, of the online generator. Your table tells whether or not $r$ should be true, as a function of the truth value of $p$ and $q$. The other shows when is the implication true. $\endgroup$ – amrsa Feb 13 '18 at 20:58
  • $\begingroup$ This might be a case of missing parentheses. Is the formula $(\lnot p \lor q)\rightarrow r$, or is it $(\lnot p) \lor (q\rightarrow r)$ ?? $\endgroup$ – hardmath Feb 13 '18 at 21:02
  • $\begingroup$ "I fail to understand how r can have different states despite p and q remaining the same" Because $r$ isnt a result of s and q. r is a third independent variable. The question isn't evaluate the truth of (Jack is not a bird) OR (mike is a horse). The question is evaluate the truth of IF [(Jack is not a bird) OR (mike is a horse) THEN (Paul is an elephant)]. $\endgroup$ – fleablood Feb 13 '18 at 22:53
1
$\begingroup$

The “$\to$” in $\sim p\lor q\to r$ does not denote a "gives", but is a logical operator, with the following truth table: $$\begin{array}{cc|c} a & b & a\to b\\ \hline 0 & 0 & 1\\ 0 & 1 & 1\\ 1 & 0 & 0\\ 1 & 1 & 1 \end{array}$$ That is, $a\to b$ is a logical statement that can be true or false. It basically says "$b$ is at least as true as $a$".

It is completely equivalent with $\sim a\lor b$.

$\endgroup$
  • $\begingroup$ So basically the rightmost column is true if the state of p q and r are consistent with the logic of the statement? $\endgroup$ – user2651000 Feb 13 '18 at 21:00
  • $\begingroup$ @user2651000: I think you could express it that way, yes. $\endgroup$ – celtschk Feb 13 '18 at 21:02
  • $\begingroup$ I would say you simply misunderstood the question. The table is evaluating a statment with three variables. $S(p,q,r)$ and you are evaluating the truth values for the 8 possible states of $p,q,r$. Think you misunderstood this as statement $S(p,q) = r$ and thought to evaluate the values of the statement with two variable for the 4 possible states of $p, q$. But that is the wrong why of interpreting the question. $\endgroup$ – fleablood Feb 13 '18 at 22:27
1
$\begingroup$

I would say you simply misunderstood the question. The table is evaluating a statment with three variables. $S(p,q,r)$ and you are evaluating the truth values for the 8 possible states of $p,q,r$. Think you misunderstood this as statement $S(p,q) = r$ and thought to evaluate the values of the statement with two variable for the 4 possible states of $p, q$. But that is the wrong why of interpreting the question.

You are being asked to evaluate the truth of $S(p,q,r) = \lnot p \lor q \to r$

which is:

$$\begin{array}{ccc|c} p&q & r & S=(\lnot p\lor q) \to r\\ \hline 0 & 0 & 0 &0\\ 0 & 0 & 1 &1\\ 0 & 1 & 0& 0\\ 0 & 1 & 1& 1\\ 1 & 0 & 0 & 1\\ 1 & 0 & 1 & 1\\ 1 & 1 & 0& 0\\ 1 & 1 & 1& 1\\ \end{array}$$

You interpreted it to mean to evaluat the truth of $R(p,q) = \lnot p \lor q$ which you evaluated as:

$$\begin{array}{cc|c} p&q & R=(\lnot p\lor q)\\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0\\ 1 & 1 & 1\\ \end{array}$$

We can combine the two results to get a clearer truth table:

$$\begin{array}{cc|c|c|c} p&q &R=\lnot p \lor q & r& S=(\lnot p\lor q) \to r=R\to r\\ \hline 0 & 0 & 1 &0&0\\ 0 & 0 & 1 &1&1\\ 0 & 1 & 1& 0&0\\ 0 & 1 & 1& 1&1\\ 1 & 0 & 0 & 0&1\\ 1 & 0 & 0 & 1&1\\ 1 & 1 & 1& 0&0\\ 1 & 1 & 1& 1&1\\ \end{array}$$

$\endgroup$
1
$\begingroup$

You're not using the online tool incorrectly; you're doing your own truth-table incorrectly.

The whole point of a truth-table is to explore all possible truth-assignments to the propositional variables involved. So, given that each of the three variables involved, $p$, $q$, and $r$ can take on the value of either $0$ or $1$, there are $2^3=8$ possible truth-assignments that the table needs to consider, i.e. your table needs $8$ rows, not $4$. The online tool has it exactly right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.