Prove $((P \cup Q) \cap (P \to R) \cap (Q \to R)) \to R $ is a Tautology $[(P \cup Q) \cap (P \rightarrow R) \cap (\neg Q \cup R)] \rightarrow R $ Logical Equivalences.
$[(P \cup Q) \cap (\neg P \cup R) \cap (\neg Q \cup R)] \rightarrow R$ Logical Equivalences.
$\neg[(P \cup Q) \cap (\neg P \cup R) \cap (\neg Q \cup R)] \cup R$ Logical Equivalences.
$[\neg(P \cup Q) \cup \neg (\neg P \cup R) \cup \neg (\neg Q \cup R)] \cup R$ De Morgan’s Laws.
$(\neg P \cap \neg Q) \cup (P \cap \neg R) \cup (Q \cap \neg R) \cup R$ De Morgan’s Laws.
I'm thinking I can use the distributive property here, but I don't know how I would set that up.
$(\neg P \cup P) \cap (\neg P \cup ¬R) \cap (\neg Q \cup P) \cap (\neg Q \cup \neg R)$ $\leftarrow$ That is as far as I have gotten as I am unsure if I continue to distribute the letters throughout the entire thing or not like this instead:
$(\neg P \cup P \cup Q \cup R) \cap (\neg P \cup \neg R \cup \neg R \cup R) \cap (\neg Q \cup P \cup Q \cup R) \cap (\neg Q \cup \neg R \cup \neg R \cup R).$
Now that I look at it, I think the second way looks correct, but I'm still unsure.
 A: Hint:
use modus ponendo ponens:
$P\land (P\implies Q)\implies Q$.
And the distributive rules.
A: $[\neg (P \cup Q) \cup \neg(\neg P \cup R) \cup \neg(\neg Q \cup R)] \cup R.$ (line 4 of your derivation)
$[\neg (P \cup Q) \cup (P \cap \neg R) \cup (Q \cap \neg R)] \cup R.$
$\{\neg (P \cup Q) \cup [(P \cup Q) \cap \neg R]\} \cup R.$
$\neg (P \cup Q) \cup \{ [(P \cup Q) \cap \neg R] \cup R\}.$
$\neg (P \cup Q) \cup \{[(P \cup Q) \cup R] \cap (\neg R \cup R)\}.$
$\neg (P \cup Q) \cup \{[(P \cup Q) \cup R] \cap \textbf{T}\}.$
$\neg (P \cup Q) \cup [(P \cup Q) \cup R].$
Can you take it from here?
A: If $R$ is true, then $\to$ is true. ($A\to T=T$), so $R$ must be false for the tautology to fail.
So we need to prove that if $R=F$ then the LHS is false.
As $A\to F=\lnot A$, from $(P\to R)\cap (Q\to R)$ we get $(P\to F)\cap (Q\to F)=\lnot P\cap\lnot Q=\lnot(P\lor Q)$
So the LHS becomes $(P\lor Q)\land\lnot(P\lor Q)=F$.
A: Truth tables are often a convenient alternative. We obtain
\begin{align*}
\begin{array}{ccccccccccccccccc}
((&P& \cup& Q&) \cap ((&P& \to &R&) &\cap& (&Q& \to& R&))) &\to& R\\
\hline
&1&1&1&1&1&1&1&&1&&1&1&1&&\color{blue}{1}&1\\
&1&1&1&0&1&0&0&&0&&1&0&0&&\color{blue}{1}&0\\
&1&1&0&1&1&1&1&&1&&0&1&1&&\color{blue}{1}&1\\
&1&1&0&0&1&0&0&&0&&0&1&0&&\color{blue}{1}&0\\
&0&1&1&1&0&1&1&&1&&1&1&1&&\color{blue}{1}&1\\
&0&1&1&0&0&1&0&&0&&1&0&0&&\color{blue}{1}&0\\
&0&0&0&0&0&1&1&&1&&0&1&1&&\color{blue}{1}&1\\
&0&0&0&0&0&1&0&&1&&0&1&0&&\color{blue}{1}&0\\
\hline
&1& 4& 2&6 &1& 4 &3& &5& &2& 4& 3& &7& 3\\
\end{array}
\end{align*}
and the claim follows.
The bottom row in the table indicates the order of evaluation.
