Assistance in completing the proof: (P → Q)∧(Q → R) is equivalent to (P → R)∧ [(P ↔ Q) ∨ (R ↔ Q)] using logical equivalencies
Solution 1:
Here are some useful but elementary equivalence principles:
Complement
$$P \lor \neg P \Leftrightarrow \top$$
$$P \land \neg P \Leftrightarrow \bot$$
Annihilation
$$P \lor \top \Leftrightarrow \top$$
$$P \land \bot \Leftrightarrow \bot$$
Identity
$$P \land \top \Leftrightarrow P$$
$$P \lor \bot \Leftrightarrow P$$
Idempotence
$$P \lor P = P$$
$$P \land P = P$$
Also, as you noticed, the whole big right terms does indeed not get you anywhere ... you need to work in the left term $\neg P \lor R$
So, starting a few lines from before you get 'stuck' (because indeed, you're just going in loops at that point) (and also throwing in some necessary parentheses):
$(\neg P \lor R) \land [\color{red}((\neg P \lor Q) \land (P \lor \neg Q)\color{red}) \lor \color{red}((\neg R \lor Q) \land (R \lor \neg Q)\color{red})] =$
$(\neg P \land [((\neg P \lor Q) \land (P \lor \neg Q)) \lor ((\neg R \lor Q) \land (R \lor \neg Q))]) \lor (R \land [((\neg P \lor Q) \land (P \lor \neg Q)) \lor ((\neg R \lor Q) \land (R \lor \neg Q))]) =$
$[\neg P \land ((\neg P \lor Q) \land (P \lor \neg Q))] \lor [\neg P \land ((\neg R \lor Q) \land (R \lor \neg Q))] \lor [R \land ((\neg P \lor Q) \land (P \lor \neg Q))] \lor [R \land ((\neg R \lor Q) \land (R \lor \neg Q))] =$
(dropping unncessary parentheses)
$[\neg P \land (\neg P \lor Q) \land (P \lor \neg Q)] \lor [\neg P \land (\neg R \lor Q) \land (R \lor \neg Q)] \lor [R \land (\neg P \lor Q) \land (P \lor \neg Q)] \lor [R \land (\neg R \lor Q) \land (R \lor \neg Q)]$
OK, now two handy laws are:
Absorption
$$P \land (P \lor Q) = P$$
Reduction
$$P \land (\neg P \lor Q) = P \land Q$$
Applying these, we get:
$[\neg P \land \neg Q] \lor [\neg P \land (\neg R \lor Q) \land (R \lor \neg Q)] \lor [R \land (\neg P \lor Q) \land (P \lor \neg Q)] \lor [R \land Q ]$
OK, and now 'unDistribute' the $\neg P $ and the $R$:
$= [\neg P \land (\neg Q \lor ((\neg R \lor Q) \land (R \lor \neg Q)))] \lor [R \land (((\neg P \lor Q) \land (P \lor \neg Q)) \lor Q) ]$
and now you can distribute the $\neg Q$ and the $Q$:
$= [\neg P \land (\neg Q \lor (\neg R \lor Q)) \land (\neg Q \lor (R \lor \neg Q))] \lor [R \land ((\neg P \lor Q) \lor Q) \land ((P \lor \neg Q) \lor Q) ] =$
(dropping unncessary parentheses)
$[\neg P \land (\neg Q \lor \neg R \lor Q) \land (\neg Q \lor R \lor \neg Q)] \lor [R \land (\neg P \lor Q \lor Q) \land (P \lor \neg Q \lor Q) ]$
And now you can use those simplification laws from the start of my post:
(Complement:)
$[\neg P \land (\neg R \lor \top) \land (R \lor \neg Q)] \lor [R \land (\neg P \lor Q) \land (P \lor \top) ]$
(Annihilation:)
$=[\neg P \land \top \land (R \lor \neg Q)] \lor [R \land (\neg P \lor Q) \land \top ]$
(Identity:)
$=[\neg P \land (R \lor \neg Q)] \lor [R \land (\neg P \lor Q)]$
(Distribution:)
$=(\neg P \land R) \lor (\neg P \land \neg Q) \lor (R \land \neg P) \lor (R \land Q)$
(Commutation:)
$=(\neg P \land R) \lor (\neg P \land \neg Q) \lor (\neg P \land R) \lor (R \land Q)$
(Idempotence:)
$=(\neg P \land R) \lor (\neg P \land \neg Q) \lor (R \land Q)$
(Distribution 2*2*2:)
$=(\neg P \lor \neg P \lor R) \land (\neg P \lor \neg Q \lor R) \land (\neg P \lor \neg P \lor Q) \land (\neg P \lor \neg Q \lor Q) \land (R \lor \neg P \lor R) \land (R \lor \neg Q \lor R) \land (R \lor \neg P \lor Q) \land (R \lor \neg Q \lor Q)$
(Complement:)
$=(\neg P \lor R) \land (\neg P \lor \neg Q \lor R) \land (\neg P \lor Q) \land (\neg P \lor \top) \land (\neg P \lor R) \land (\neg Q \lor R) \land (R \lor \neg P \lor Q) \land (R \lor \top)$
(Annihilation:)
$=(\neg P \lor R) \land (\neg P \lor \neg Q \lor R) \land (\neg P \lor Q) \land \top \land (\neg P \lor R) \land (\neg Q \lor R) \land (R \lor \neg P \lor Q) \land \top$
(Identity:)
$=(\neg P \lor R) \land (\neg P \lor \neg Q \lor R) \land (\neg P \lor Q) \land (\neg P \lor R) \land (\neg Q \lor R) \land (R \lor \neg P \lor Q) $
(two Absorptions and an Idempotence:)
$=(\neg P \lor R) \land (\neg P \lor Q) \land (\neg Q \lor R)$
Phew! Almost there ....
Now, use:
Adjacency
$$P = (P \lor Q) \land (P \lor \neg Q)$$
Applied to where we were:
$(\neg P \lor R) \land (\neg P \lor Q) \land (\neg Q \lor R)$
(Adjacency:)
$=(\neg P \lor R \lor Q) \land (\neg P \lor R \lor \neg Q) \land (\neg P \lor Q) \land (\neg Q \lor R)$
(Two Absorptions)
$(\neg P \lor Q) \land (\neg Q \lor R)$
.. and finally we're there! Sheesh!