# Truth table confusion

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?

• And what is the question you've been asked? – amrsa Feb 13 '18 at 20:52
• @amrsa To write a truth table for the conditional statement "~p ∨ q -> r" – user2651000 Feb 13 '18 at 20:55
• 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. – amrsa Feb 13 '18 at 20:58
• This might be a case of missing parentheses. Is the formula $(\lnot p \lor q)\rightarrow r$, or is it $(\lnot p) \lor (q\rightarrow r)$ ?? – hardmath Feb 13 '18 at 21:02
• "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)]. – fleablood Feb 13 '18 at 22:53

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 $\sim a\lor b$.

• So basically the rightmost column is true if the state of p q and r are consistent with the logic of the statement? – user2651000 Feb 13 '18 at 21:00
• @user2651000: I think you could express it that way, yes. – celtschk Feb 13 '18 at 21:02
• 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. – fleablood Feb 13 '18 at 22:27

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:

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

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.