1
$\begingroup$

I tried to use truth tables but couldn't get it, I used logical equivalences and got it, can someone show me how to do it with a truth table?

$\endgroup$
  • 2
    $\begingroup$ How did you use equivalences and 'got it' when they are not equivalent? $\endgroup$ – Bram28 Jan 24 at 2:17
0
$\begingroup$

You probably just miscalculated the truth table. Here is the complete truth table for the two statements:

$$\begin{array}{ccc|c|c} p & q & r & p\wedge q \to r & p\vee q\to r \\ \hline T & T & T & T & T \\ T & T & F & F & F \\ T & F & T & T & T \\ T & F & F & T & F \\ F & T & T & T & T \\ F & T & F & T & F \\ F & F & T & T & T \\ F & F & F & T & T \end{array} $$

It is easy to see that the two statements are not equivalent.

An easy way to work out the truth table for a statement of the form $A\to B$ is to note that this will only be false when $A$ is true but $B$ is false, and is true everywhere else. This fact is what I used to calculate the tables above.

$\endgroup$
2
$\begingroup$

$p=1,q=0,r=0, p\wedge q = 0, p\vee q=1, p\wedge q\rightarrow r = 1, p\vee q\rightarrow r = 0$

$\endgroup$
0
$\begingroup$

\begin{array}{ccc|ccc|ccc} p&q&r&(p \land q)& \rightarrow &r&(p\lor q)& \rightarrow &r\\ \hline T&T&T&T&T&T&T&T&T\\ T&T&F&T&F&F&T&F&F\\ T&F&T&F&T&T&T&T&T\\ T&F&F&F&\color{red}T&F&T&\color{red}F&F\\ F&T&T&F&T&T&T&T&T\\ F&T&F&F&\color{red}T&F&T&\color{red}F&F\\ F&F&T&F&T&T&F&T&T\\ F&F&F&F&T&F&F&T&F\\ \end{array}

$\endgroup$
0
$\begingroup$

can someone show me how to do it with a truth table?

If and only if they were logically equivalent, then you could construct a table such that $p\land q\to r$ and $p\lor q\to r$ are given identical truth values by each assignment of $p,q,$ and $r$.

So you would have to construct a table for the eight assignments of the literals and verify that that happens.

However, that can be a lot of effort.   We only need one counterexample to show that they are not equivalent.   That is, one assignment of $p,q,r$ which gives differing values for the conditionals.

So if we could see a way for this to happen, then it would save all that work.   Now how might that happen?

Why now, a conditional is only falsified when the antecedent is true while the consequent is false.   So let us assign $r$ to be false and think about which evaluations of $p,q$ are needed to make one of $p\land q$ or $p\lor q$ false while the other is true (falsifying one conditional but not the other).

$\endgroup$

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.