# Simplify/Show Boolean Logic expression is a Tautology

I need to prove that the expression $$(p\leftrightarrow q) \wedge (q\leftrightarrow r) \rightarrow (p\leftrightarrow r)$$ is a tautology, I have managed to simplify it down to the expression $$pr+p'r'+pq'+p'q+rq'+r'q$$ or alternatively $$(p\land r)\lor(\lnot p\land \lnot r) \lor (p\land \lnot q)\lor(\lnot p\land q)\lor(r\land \lnot q)\lor(\lnot r\land q)$$ . But I haven't been able to show that it is a tautology without using a truth table. How can I simplify it further?

• What do the dashes in the superscript mean? Commented Jan 30, 2022 at 19:23
• Thanks for pointing it out, they mean complement. I'll edit it to remove the dashes from the last expression Commented Jan 30, 2022 at 19:24
• Why don't you use the table of truth? Commented Jan 30, 2022 at 19:29
• I agree. I tried simplifying it. But that probably just makes it more difficult or complex. Commented Jan 30, 2022 at 19:33
• I verified it using a truth table but I just want to try and prove it without a truth table Commented Jan 30, 2022 at 19:33

Note that $$p = p1 = p(q+q')=pq+pq'$$

We can use this to 'split' all your $$6$$ terms:

$$pr+p'r'+pq'+p'q+rq'+r'q=$$

$$prq+prq'+p'r'q+p'r'q'+pq'r+pq'r'+p'qr+p'qr'+rq'p+rq'p'+r'qp+r'qp'$$

Reorder each term alphabetically:

$$pqr+pq'r+p'qr'+p'q'r'+pq'r+pq'r'+p'qr+p'qr'+pq'r+p'q'r+pqr'+p'qr'$$

And reorder the 12 terms:

$$pqr+pqr'+pq'r+pq'r+pq'r+pq'r'+p'qr+p'qr'+p'qr'+p'qr'+p'q'r+p'q'r'$$

Remove duplicates:

$$pqr+pqr'+pq'r+pq'r'+p'qr+p'qr'+p'q'r+p'q'r'$$

And put them back together:

$$pqr+pqr'+pq'r+pq'r'+p'qr+p'qr'+p'q'r+p'q'r'=$$

$$pq+pq'+p'q+p'q'=$$

$$p+p'=$$

$$1$$