# If $(P \lor Q) = (P \lor R)$ Can we conclude $Q = R$ ? What happens if we use AND instead of the operator OR?

So I am trying to solve this problem and not quite able to figure out the answer. So I'm not sure what two statements I could construct in my truth table to prove these true or false. I believe that $Q$ is not equal to $R$ because OR statements are inclusive. And I believe if we used the AND operators they would be equal. The statement I was thinking about constructing to prove this was if $R \land (P \lor Q) / Q \land (P \lor R)$? Any help would be appreciated

• Doesn't follow, whether you use And or Or. For the Or case, consider the truth-assignment $P \equiv R \equiv \top$ and $Q \equiv \bot$. For the second, consider $P \equiv Q \equiv \bot$ and $R \equiv \top$. May 6, 2014 at 8:42
• @HunanRostomyan You could post that as an answer. May 6, 2014 at 14:08
• Could we say that if we assume Q and R to both be true that when we put them into the "AND" statements it means that P would always have to be true therefore they are not equivalent? Then for the "OR" statements if we assume both Q and R are false then if (P OR Q) is going to be equal to (P OR R) P would always have to be false which we know is also not the case therefore both statements show that Q != R May 6, 2014 at 15:55

If P is true, both P OR Q and P OR R are both true regardless of what Q and R may be. Similarly for the AND case, both are false if P is false.

• @DougSpoonwood How does this contradict what I said? I showed cases where both can be equal without $Q=R$. Which happens to be what the question is about.
– Mike
May 6, 2014 at 16:20