[Math] Proving $[(p\leftrightarrow q)\land(q\leftrightarrow r)]\to(p\leftrightarrow r)$ is a tautology without a truth table

logicpropositional-calculus

I came across the following problem in a book:

Show that if $p, q$, and $r$ are compound propositions such that $p$ and $q$ are logically equivalent and $q$ and $r$ are logically equivalent, then $p$ and $r$ are logically equivalent.

The book's solution certainly makes sense:

To say that $p$ and $q$ are logically equivalent is to say that the truth tables for $p$ and $q$ are identical; similarly, to say that $q$ and $r$ are logically equivalent is to say that the truth tables for $q$ and $r$ are identical. Clearly if the truth tables for $p$ and $q$ are identical, and the truth tables for $q$ and $r$ are identical, then the truth tables for $p$ and $r$ are identical. Therefore $p$ and $r$ are logically equivalent.

I decided to "symbolically translate" the problem in the book:

Show that $[(p\leftrightarrow q)\land(q\leftrightarrow r)]\to(p\leftrightarrow r)$ is a tautology.

I wrote out a truth table and everything checks out, as expected (and as mentioned in the book's solution). My question is whether or not there is a more "algebraic" solution using equivalences (not resorting to CNF or DNF).

Any ideas?

Best Answer

I managed to work out a solution, and I thought I would share even though it is rather hideous (picture posted to make formatting much more pleasant to read). enter image description here

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