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

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 $R$.



Daniel Clemente Laboreo 2005-05-17