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So far, this is what I have done:

Let LHS = (p ∨ q) ∨ (p ∧ r) and RHS = (p ∨ q) ∧ r

LHS =((p ∨ q) ∨ p) ∧ ((p ∨ q) ∨ r)

= (p ∨ q) ∧ ((p ∨ q) ∨ r)

= (p ∨ q) ∧ (p ∨ q ∨ r)

= (p ∨ q) ∧ (p ∨ q) ∨ r

= (p ∨ q) ∨ r

Hence, (p ∨ q) ∨ r ̸≡ (p ∨ q) ∧ r as they have opposite truth values. i.e. when p and q = F and r = T.

May I ask if such a presentation is considered as correct?

4 Answers 4

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I think you could just give a counter example. Let $p$ and $q$ be true and $r$ be false.

Then the left hand side is true and the right hand side is false.

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Really, the only part you need is the last step - find a truth assignment which yields different values for each statement. You can do this right at the beginning.

However, you've made a mistake in your analysis of the LHS: plug $p=F, q=F, r=T$ into the original LHS and you'll see that it's false. Your error is when you mix up the parentheses in your last line: $(p \vee q) \wedge ((p \vee q) \vee r)$ is not the same as $((p\vee q)\wedge(p\vee q))\vee r$. You're assuming that $\vee$ and $\wedge$ associate through each other, but this isn't true - e.g. $(A\vee B)\wedge C$ is not $A\vee (B\wedge C)$ in general. In fact, the LHS is equivalent to just $p\vee q$ (do you see why?).

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    I am able to see that $(p \vee q) \vee p)$ equivalent to $p\vee q$ but I wasn't able to see how $((p \vee q) \vee r)$ is equivalent to $p\vee q$. Would you mind elaborating more? Further on this, I still couldn't identify where my error lies as I have used distributive law for my first rewriting of LHS. – 2017-02-19
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    @WeitingChen Sorry, I misread what you wrote - indeed you've applied the distributive law correctly. Your error is when you mix up the parentheses in your last line: $(p \vee q) \wedge ((p \vee q) \vee r)$ is not the same as $((p\vee q)\wedge(p\vee q))\vee r$. You're assuming that $\vee$ and $\wedge$ associate through each other, but this isn't true - e.g. $(A\vee B)\wedge C$ is not $A\vee (B\wedge C)$ in general. In fact, the LHS is equivalent to just $p\vee q$ (do you see why?). – 2017-02-19
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To easily come up with a counterexample (instead of just randomly assigning truth values and hoping for one), I would use distributivity to note that $$ (p\lor q)\land r\equiv(q\land r)\lor(p\land r).\tag{1} $$ Since you are comparing $(1)$ to $$ (p\lor q)\lor(p\land r),\tag{2} $$ it is pretty easy to note that you can simply let $r$ be false, whereby $(1)$ is clearly false, but $(2)$ depends on whether or not $p$ or $q$ (or both) is true. Simply let one or both of them be true to obtain a counterexample.

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$$\begin{align} (p \lor q) \lor (p \land r) &\equiv (p \lor q\lor p) \land (p\lor q \lor r)\\ \\ &\equiv (p \lor q) \land (p\lor q\lor r)\\ \\ &\equiv (\underbrace{(p\lor q)\land (p\lor q))}_{ \large = p\lor q}\lor {((p \lor q)\land r)}\tag{Distibution}\\ \\ &\equiv (p\lor q) \end{align}$$

Now note that if $p$ or $q$ is true, this left hand side is true, and the truth value of $r$ doesn't matter. In contrast, for the RHS $(p\lor q) \land r$ is false whenever $r$ is false, even if $p\lor q$ is true.