Prove that $(p \to q) \land (q \to r)$ is equivalent to $p \to r$

Question

$(P\implies Q)\land(Q\implies R)$ is equivalent to $P\implies R$. Is this true? How to prove this directly, not using truth tables?

Answer 1

If $P$ and $R$ are true, then $P \implies R$ is true no matter what $Q$ is. So just choose a $Q$ to make one of the statements $P \implies Q$ and $Q \implies R$ become false, then the two sides don't agree. In this case choosing $Q$ to be false works.

Answer 2

They aren’t equivalent. You mentioned wanting a proof without truth tables, but here is one, anyway. $\newcommand{\tru}{{\color{#0c0}T}}$ $\newcommand{\fal}{{\color{#c00}F}}$ \begin{array}{|c|c|c|c|c|} \hline P & Q & R & (P \Rightarrow Q) \wedge (Q \Rightarrow R) & P \Rightarrow R \\ \hline \fal & \fal & \fal & \tru & \tru \\ \fal & \fal & \tru & \tru & \tru \\ \fal & \tru & \fal & \fal & \tru \\ \fal & \tru & \tru & \tru & \tru \\ \tru & \fal & \fal & \fal & \fal \\ \tru & \fal & \tru & \fal & \tru \\ \tru & \tru & \fal & \fal & \fal \\ \tru & \tru & \tru & \tru & \tru \\ \hline \end{array}

It is true, however, that

$$(P \Rightarrow Q) \wedge (Q \Rightarrow R) \text{ implies } P \Rightarrow R.$$

Answer 3

$\begin{align} & \underline{\begin{align} & P\to Q\Leftrightarrow \,\sim P\vee Q \\ & Q\to R\Leftrightarrow \,\sim Q\vee R \\ \end{align}} \\ & (P\to Q)\wedge (Q\to R)\Leftrightarrow (\sim P\vee Q)\wedge (\sim Q\vee R) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow [(\sim P\vee Q)\wedge \sim Q]\vee [(\sim P\vee Q)\wedge R] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow [(\sim P\wedge \sim Q)\vee \underbrace{(Q\,\wedge \sim Q)}_{F}]\vee [(\sim P\wedge R)\vee (Q\,\wedge R)] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow (\sim P\wedge \sim Q)\vee (\sim P\wedge R)\vee (Q\,\wedge R) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \,\,\sim P\wedge (\sim Q\vee R)\vee (Q\,\wedge R) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \,\,\sim P\wedge [(\sim Q\vee R)\vee (Q\,\wedge R)] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \,\,\sim P\wedge [(\sim Q\vee R)\vee (Q\,\wedge R)] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \,\,\sim P\wedge (\,\underbrace{[(\sim Q\vee R)\vee Q\,]}_{R}\wedge \underbrace{[(\sim Q\vee R)\vee R}_{\sim Q\vee R}\,]) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \to \,\,\sim P\wedge R \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \to \,\,P\to R \\ \end{align}$