5.2 A bit more complicated. $P\wedge Q\Rightarrow R,\ Q\Rightarrow P,\ Q\vdash R$

Try yourself $P\wedge Q\Rightarrow R,\ Q\Rightarrow P,\ Q\vdash R$. Then look the solution:

$$\begin{fitch} \par P \wedge Q \Rightarrow R \\ \par Q \Right... ...\wedge$\ 4,3 \\ \par R & E$\Rightarrow$\ 1,5 \par \end{fitch}$$

The only way to achieve $R$ is using the first formula, $P\wedge Q\Rightarrow R$, but we can only use it when $P\wedge Q$ is true, so we're going for that.

We know that $Q\Rightarrow P$ (line 2) and also $Q$ (line 3), so we deduce $P$. Since $P$ is now true and also $Q$, $P\wedge Q$ is too. Until now it's similar to the previous exercise.

Finally, we have $P\wedge Q\Rightarrow R$, and know that $P\wedge Q$, so we finish by saying $R$.