Discrete Math - Logic - Implication Problem I am taking a discrete mathematics course, and I am currently studying the logic chapters of the textbook. The problem I am trying to solve at the moment says "Use truth tables to verify the following" which makes me assume that it expects the two compound statements being compared to be logically equivalent. Anyways this is the problem:
$$
(P\implies Q)\land(Q\implies R)\equiv (P\implies R).
$$
When I set up the truth table for this problem, I am getting the answer that thee two statements are not logically equivalent. Perhaps this is right, but this problem has 4 other subproblems (a,b,c,d,....etc.), each of which had statements that were logically equivalent. Maybe I am over-thinking this, but any help, hint, or explanation is appreciated.
 A: You are correct that $$(P \rightarrow Q) \land  (Q \rightarrow R) \not\equiv P\rightarrow R$$
It is true that $$\Big((P\rightarrow Q) \land (Q\rightarrow R)\Big) \rightarrow (P \rightarrow R)$$ (You can confirm that this is a tautology by consulting its truth-table.)
But if $Q$ is false, and P is true, and R is true, the left hand side is false, but the righthand side is true, so the implication is not bidirectional.
A: I think you're meant to be proving $$\big((P \implies Q) \wedge (Q \implies R)\big) \implies (P \implies R)$$ is a tautology.  This is true, as per the following truth table:
$$
\begin{array}{|ccc|ccc|c|c|}
\hline
P & Q & R & P \implies Q & Q \implies R & (P \implies Q) \wedge (Q \implies R) & P \implies R & \text{above} \\
\hline
0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}
$$
Comparing the last two columns, we see the identity holds.
A: The statement above is true, because the implication is false only if the premise is true and the conclusion is false.
