# Show that p ∧ q → r and p ∨ q → r are not logically equivalent.

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?

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

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.

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

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

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).