Simplify this propositional statement

Question

how can I simplify $(p\to q)\wedge(q\to r)$?

I need to find out for which $p, q, r$ this is true, without using truth tables.

Answer 1

Using $p\rightarrow q\equiv \neg p\vee q$ and the distributive laws, convert $(p\rightarrow q)\wedge(q\rightarrow r)$ into its DNF:

$$(p\rightarrow q)\wedge(q\rightarrow r)\equiv (\neg q\wedge \neg p)\vee (r\wedge\neg p)\vee (r\wedge q)$$

Since this is a disjunction of three clauses, it is true if any of the three clauses are true.

For example, the first clause is true if and only if both $p$ and $q$ are false irrespective of the truth value of $r$. So the truth values $FFT, FFF$ of $p, q, r$ makes the expression true. Similarly for second and third clauses will give that $FTT$ and $TTT$ are the only other possibilities.

Answer 2

$(p \to q) \land (q \to r) \Leftrightarrow $ (Implication)

$(\neg p \lor q) \land (\neg q \lor r) \Leftrightarrow$ (Distribution)

$(\neg p \land \neg q) \lor (\neg p \land r) \lor (q \land \neg q) \lor (q \land r) \Leftrightarrow $ (Complement)

$( \neg p \land \neg q) \lor (\neg p \land r) \lor \bot \lor (q \land r) \Leftrightarrow$ (Identity)

$(\neg p \land \neg q) \lor (\neg p \land r) \lor (q \land r) \Leftrightarrow $ (Adjacency)

$(\neg p \land \neg q \land r) \lor (\neg p \land \neg q \land \neg r) \lor (\neg p \land q \land r) \lor (\neg p \land \neg q \land r) \lor (p \land q \land r) \lor (\neg p \land q \land r) \Leftrightarrow$ (Idempotence)

$(\neg p \land \neg q \land r) \lor (\neg p \land \neg q \land \neg r) \lor (\neg p \land q \land r) \lor (p \land q \land r)$

And now you have an explicit expression for all the different combinations of $p$, $q$, and $r$ for which this statement is true.