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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?

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    $\begingroup$ How did you use equivalences and 'got it' when they are not equivalent? $\endgroup$
    – Bram28
    Jan 24, 2019 at 2:17

4 Answers 4

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

Edit: To figure out counterexamples such as this it is often instructive to rewrite the implication in terms of disjunction or conjuction, and negation:

$$a\to b \equiv \lnot a \lor b$$

Then:

$$(p\lor q\to r)\land(p\land q\to r) \equiv \left(\lnot(p\lor q)\lor r\right)\land(\lnot(p\land q)\lor r)$$

From here you figure out that if $r=1$ then the above is always true, so you choose $r=0$ and try to find $p,q$ for which $\lnot(p\lor q)\ne \lnot(p\land q)$, which is the same as: $p\lor q\ne p\land q$. The latter occurs when $p\ne q$, then you pick $p=0, q=1$ or $p=1, q=0$.

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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.

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\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}

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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).

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