Need help with solving proposition logic formula, should be a tautology I have the following formula:
$(((p \vee q) \rightarrow r) \wedge (p \rightarrow q))\rightarrow (q\rightarrow r)$
The truth table for this formula shows that this is a tautology. However, I get stuck simplifying this formula at a certain point. My steps are as follows:
First, I use implication elimination on all $\rightarrow$. The formula will then become:
$\neg ((\neg(p \vee q) \vee r) \wedge (\neg p \vee q)) \vee (\neg q\vee r)$
Then, I will use De Morgan on $\neg(p \vee q)$:
$\neg (((\neg p \wedge \neg q) \vee r) \wedge (\neg p \vee q)) \vee (\neg q \vee r)$
And then De Morgan on $\neg (((\neg p \wedge \neg q) \vee r) \wedge (\neg p \vee q))$:
$(\neg ((\neg p \wedge \neg q) \vee r) \vee \neg (\neg p \vee q)) \vee (\neg q \vee r)$
$((\neg (\neg p \wedge \neg q) \wedge \neg r) \vee \neg (\neg p \vee q)) \vee (\neg q \vee r)$
$(((\neg \neg p \vee \neg \neg q) \wedge \neg r) \vee \neg (\neg p \vee q)) \vee (\neg q\vee r)$
$(((p \vee q) \wedge \neg r) \vee \neg (\neg p \vee q)) \vee (\neg q \vee r)$
$(((p \vee q) \wedge \neg r) \vee (\neg \neg p \wedge \neg q)) \vee (\neg q \vee r)$
$(((p \vee q) \wedge \neg r) \vee (p \wedge \neg q)) \vee (\neg q \vee r)$
From here I have no ideas anymore what to do. All things seem to go into a dead end. For example, I tried to distribute $\neg r$ over $(p \vee q)$ in $((p \vee q) \wedge \neg r)$, and then I'd end up with:
$((\neg r \wedge p) \vee (\neg r \wedge q) \vee (p \wedge \neg q)) \vee (\neg q \vee r)$
With this, I also have no idea what to do.
It would be great if I'd get some help with how to proceed from here, how to get to the tautology.
 A: I'll go rather slowly, but I will omit some superfluous parentheses.
Denoting $(((p \vee q) \rightarrow r) \wedge (p \rightarrow q))\rightarrow (q \rightarrow r)$ by $(\star)$ we have
$$\begin{align}
(\star )
%%
&=
\neg ( ( \neg ( p \vee q ) \vee r ) \wedge ( \neg p \vee q ) ) \vee ( \neg q \vee r ) &\text{(unabbreviating)} \\
&= ( \neg ( \neg ( p \vee q ) \vee r ) \vee \neg ( \neg p \vee q ) ) \vee ( \neg q \vee r ) &\text{(de Morgan)} \\
%%
&= ( ( \neg \neg ( p \vee q ) \wedge \neg r ) \vee ( \neg \neg p \wedge \neg q ) ) \vee ( \neg q \vee r ) &\text{(de Morgan)} \\ 
%%
&= ( ( p \vee q ) \wedge \neg r ) \vee ( p \wedge \neg q ) \vee ( \neg q \wedge r ) &\text{(double negation)} \\
%%
&= ( p \wedge \neg r ) \vee \color{red}{( q \wedge \neg r )} \vee ( p \wedge \neg q ) \vee \color{red}{( \neg q \vee r )} &\text{(distributivity)}\end{align}$$
Note that $$\begin{align}
\neg q \vee r &= \neg \neg \neg q \vee \neg \neg r &\text{(double negation)} \\
&= \neg ( \neg \neg q \wedge \neg r ) &\text{(de Morgan)} \\
&= \neg ( q \wedge \neg r ) &\text{(double negation)}
\end{align}$$
and so we have
$$\begin{align}
(\star) 
&= ( p \wedge \neg r ) \vee \color{red}{( q \wedge \neg r )} \vee ( p \wedge \neg q ) \vee \color{red}{( \neg q \vee r )} \\
&= 
( p \wedge \neg r ) \vee \color{red}{( q \wedge \neg r )} \vee ( p \wedge \neg q ) \vee \color{red}{\neg ( q \wedge \neg r )} &\text{(by above)}  \\ 
&= \top \vee ( p \wedge \neg r ) \vee ( p \wedge \neg q ) &\text{($s \vee \neg s = \top$)} \\
&= \top &\text{($\top \vee s = \top$)}
\end{align}$$
A: $$(((p \vee q) \rightarrow r) \wedge (p \rightarrow q))\rightarrow (p\rightarrow r)$$
is implied by
$$((p \vee q) \rightarrow r)\rightarrow (p\rightarrow r)$$
is equivalent to
$$p \rightarrow (p \vee q)$$
is implied by
$$p \rightarrow p$$
is implied by
$$\top$$
