Prove the following tautology: $\big[(p\leftrightarrow q) \land (\lnot q \to r) \land (p \to r)\big]\to r $ 
Prove the following tautology:
$$\big[(p\leftrightarrow q) \land (\lnot q \to r) \land (p \to
 r)\big]\to r $$

My effort
I am trying to prove this with a direct reasoning,i.e without using truth tables.
Now since $ \lnot q \to r $ and $p \to r $ ,I think it follows that I can rewrite $$\big[(p\leftrightarrow q) \land (\lnot q \to r) \land (p \to
 r)\big]=\big[(p \land \lnot q) \to r \big] \to r $$
which would be the proof for it,but I am not so sure :I just started studying  math logic.
In what other ways (excluding truth table) could this problem be solved  ?
 A: I think there's a good direct reasoning method with a different starting point. Reminder that we need to confirm that if all the statements on the LHS are true, then the statement on the RHS is true. So we will assume the LHS is true and see if that forces $r$ to hold.
Consider $p\iff q$. Then either both are true or both are false. If both are true, then since $p\Rightarrow r$, we have that $r$ holds. If both are false, then since $\neg q \Rightarrow r$, we have that $r$ holds. Thus the entire expression implies $r$.
So why does this prove what we are interested in? I am establishing that, given the truth of the LHS, it follows that $r$ must hold. Put another way, if $\neg r$ was to hold, it would be contradictory. If we recall the definition of $A\Rightarrow B$, we see that $A\Rightarrow B$ holds if and only if $\neg A$ holds or ($A$ holds and $B$ holds). 
You never have to worry about the $\neg A$ case, because the relationship between $A$ and $B$ is irrelevant. If $\neg A$ is possible, then the implication holds for values that make it so. If $\neg A$ never holds (so $A$ is a tautology) then that case of the implication is trivially or vacuously true.
So, we need to concern ourselves with both $A$ and $B$ holding. To this end, I showed that if $A$ holds, it logically follows that $B$ holds. Thus for $A$ to hold but not $B$ would be a logical contradiction, as we desire.
A: Hint:
$$\small\begin{align}
\big[(p\leftrightarrow q) \land (\lnot\, q \to r) \land (p \to
 r)\big]\to r
&
\equiv
\big[(p\to q)\land(q\to p)\land (\lnot\,(\lnot\, q) \lor r) \land (\lnot\, p \lor 
r)\big]\to r \\
&
\equiv
\big[(\lnot\, p\lor q)\land(\lnot\, q\lor p)\land (\lnot\,(\lnot\, q) \lor r) \land (\lnot\, p \lor 
r)\big]\to r \\
&
\equiv
\big[(\lnot\, p\lor q)\land(\lnot\, q\lor p)\land (q \lor r) \land (\lnot\, p \lor 
r)\big]\to r \\
&
\equiv
\big[(\lnot\, p\lor q)\land(\lnot\, p\lor q)\land r \land (p \lor\lnot\, 
p)\big]\to r \\
&
\equiv
\big[(\lnot\, p\lor q)\land(\lnot\, p\lor q)\land r\big]\to r. \\
\end{align}$$
Is $\small[\,\mathrm{P}(p,q)\land r\,]\to r$ a tautology? Can you proceed further?
A: A natural deduction proof:


*

*Assume (p↔q)∧(¬q→r)∧(p→r)

*Infer (p↔q),(¬q→r), and (p→r) by conjunction elimination 

*Assume q, infer r, discharge to get (q→r)

*You now have the consequences (¬q→r) and (q→r)

*Infer r by disjunction elimination from (p or ¬p)

*Discharge

A: A proof by resolution


*

*Start from the corresponding  set of clauses $\{(¬p\vee q),(¬q\vee p),(q\vee r),( \neg p \vee r)\}$

*Resolve clauses 2 and 4 to produce clause 5:  $(¬q \vee r)$

*Resolve  clause 3 and 5 to get (r)

A: Here is a natural deduction proof in a Fitch-style proof checker:

The law of the excluded middle (LEM) on $Q \lor \neg Q$ is referenced at the end.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
