Let $p$, $q$, and $r$ be any propositions. Then, using a chain of logical equivalences, how to establish the following logical identity? $$ \big( (p \rightarrow q) \land (q \rightarrow r) \big) \rightarrow \big( p \rightarrow r \big) \ \equiv \ T. $$
My Attempt:
In the following chain of equivalences, I will be using the same nomenclature as that used in Table 6, Sec. 1.3, of the book Discrete Mathematics and Its Applications by Kenneth H. Rosen, 8th edition.
We note that $$ \begin{align} &\qquad \big( (p \rightarrow q) \land (q \rightarrow r) \big) \rightarrow \big( p \rightarrow r \big) \\ &\equiv \left( \overline{ \left( \overline{p} \lor q \right) \land \left( \overline{q} \lor r \right) } \right) \lor \left( \overline{p} \lor r \right) \\ & \qquad \mbox{[ using the conditional-disjunction equivalence ]} \\ &\equiv \left( \left( \overline{ \overline{p} \lor q } \right) \lor \left( \overline{ \overline{q} \lor r } \right) \right) \lor \left( \overline{p} \lor r \right) \\ &\qquad \mbox{[ using a DeMorgan's law ]} \\ &\equiv \left( \left( \overline{ \overline{p}} \land \overline{q} \right) \lor \left( \overline{ \overline{q}} \land \overline{r} \right) \right) \lor \left( \overline{p} \lor r \right) \\ &\qquad \mbox{[ using a DeMorgan's law ]} \\ &\equiv \left( \left( p \land \overline{q} \right) \lor \left( q \land \overline{r} \right) \right) \lor \left( \overline{p} \lor r \right) \\ &\qquad \mbox{[ using the double-negation law ]} \\ &\equiv \left( \left( p \land \overline{q} \right) \lor \overline{p} \right) \ \lor \ \left( \left( q \land \overline{r} \right) \lor r \right) \\ &\qquad \mbox{[ using the associativity and commutativity of $\lor$ ]} \\ &\equiv \left( \left( p \lor \overline{p} \right) \land \left( \overline{q} \lor \overline{p} \right) \right) \ \lor \ \left( \left( q \lor r \right) \land \left( \overline{r} \lor r \right) \right) \\ &\qquad \mbox{[ using the distributivity of $\lor$ over $\land$ ]} \\ &\equiv \left( T \land \left( \overline{q} \lor \overline{p} \right) \right) \ \lor \ \left( \left( q \lor r \right) \land T \right) \\ &\qquad \mbox{[ using a negation law ]} \\ &\equiv \left( \overline{q} \lor \overline{p} \right) \ \lor \ \left( q \lor r \right) \\ &\qquad \mbox{[ using an identity law ]} \\ &\equiv \left( \overline{q} \lor q \right)\ \lor \ \left( \overline{p} \lor r \right) \\ &\qquad \mbox{[ using the associativity and commutativity of $\lor$ ]} \\ &\equiv T \ \lor \ \left( \overline{p} \lor r \right) \\ &\qquad \mbox{[ using a negation law ]} \\ &\equiv T. \\ &\qquad \mbox{[ using a domination law ]} \end{align} $$
Is the above calculation correct and clear enough? If so, is this the shortest possible chain of logical equivalences necessary for establishing the tautology in question, using the same machinary as developed by Rosen up to Sec. 1.3?
Or, are there any issues?
I'm with Dan Christensen. It is not only just horribly messy to use "chains of logical equivalence" it completely obscures why $((p \rightarrow q) \land (q \rightarrow r)) \rightarrow \big( p \rightarrow r)$ is logically true on any sensible treatment of the conditional (including e.g. in constructive frameworks where De Morgan's Laws, for example, aren't available).
It is worth seeing this, and as Dan says, a natural deduction framework makes this absolutely obvious:
$\quad \quad | \quad \quad (p \to q) \land (q \to r)\quad\quad temporary\ assumption\\ \quad \quad | \quad \quad (p \to q)\quad\quad\quad\quad\quad unpacking\ the\ conjunction\\ \quad \quad | \quad \quad (q \to r)\\ \quad \quad | \quad \quad | \quad \quad p\quad \quad\quad\quad\quad another\ temporary\ assumption\\ \quad \quad | \quad \quad | \quad \quad q\quad \quad\quad\quad\quad modus\ ponens!\\ \quad \quad | \quad \quad | \quad \quad r\quad \quad\quad\quad\quad modus\ ponens!\\ \quad \quad | \quad \quad (p \to r)\quad\quad\quad\quad\quad we've\ shown\ r\ follows\ from\ p\\ ((p \to q) \land (q \to r)) \to (p \to r)\quad we've\ shown\ (p \to r)\ follows\ from\ (p \to q) \land (q \to r) $
That's entirely intuitive: if we are given both $p \to q$ and $q \to r$ then obviously the supposition that $p$ gets you to $r$, so $p \to r$ -- and that's just using basic principles about the conditional (and unpacking the conjunction). There's nothing here that depends on the conditional-disjunction equivalence (not valid in e.g. intuitionistic logic) or on your use of De Morgan's laws (ditto). So I'd say that the use of "use chains of logical equivalence" is to be deprecated as hiding what is going.
Of course this is not intended as any criticism of the OP, assuming Rosen's book has been thrust upon them!