# Showing Inconsistency in Monadic Predicate Logic

Any help would really be greatly appreciated. I've attached my workings but I'm not too sure if what I've done is right or not. Thank you ever so much in advanced!

Show that the set $S=\{s_1,s_2\}$ is inconsistent by forming the sentences $s_i$ as premises and deriving a contradiction. $$S =\{ \forall x (Px \rightarrow Qx)\rightarrow\forall x (Px\rightarrow \neg Rx) , \exists x(Px \land Rx)\land\forall x(Px\rightarrow Qx)\}$$

My ideas:

Your proof is largely correct, but there are a number of minor things that need to be corrected, and in which the proof can be shortened quite a bit:

Lines 3 and 4: I am not sure about your rule, which looks like $\top I$ ... This should be $\land E$ or Simplification

Line 7: as we will see, there is no need to start a subproof here. And you already have this line on line 4. So this line can be removed

Line 8 can be removed as well

Line 9: again, no subproof necessary. Instead, get this statement from 1 and 4 through $\rightarrow E$, or Modus Ponens.

Line 11: first you need to isolate $Pa$ from 5 before you can get $\neg Ra$.

Line 12: again, first isolate $Ra$ by Simplication or $\land E$ from 5 before conjuncting $Ra$ and $\neg Ra$

... And line12 is your contradiction, hence your last line. So no more lines needed after this ... Which is why you don't need any of those subproofs.

• @Andy you're welcome! If you want, you can add a picture of your edited proof and I can take another look. – Bram28 Jan 22 '17 at 15:31
• Thank you! I will do so soon and post it here! Thanks again! – Andy Jan 22 '17 at 15:43
• @Andy OK. Make sure to leave your original proof as well so my answer still makes sense and others can learn from this post. – Bram28 Jan 22 '17 at 15:47