# Prove that the following argument is valid

I'm asked to show that the following argument is valid:

P1) $$[E \lor (L \lor M)] \land (E \leftrightarrow F)$$

P2) $$L \rightarrow D$$

P3) $$D \rightarrow \neg L$$

C) $$E \lor M$$

Here is my work (so far):

P2) $$L \rightarrow D$$

1. $$\neg(\neg L) \rightarrow D$$ Premise
2. $$L$$ Premise
3. $$L \rightarrow D$$ 1, Substitution
4. $$D$$ 2, 3 Modus

I'm not sure.

I know you need to use the rules of inference like modus ponens or converse fallacy, but I'm confused because it doesn't look like any of the forms I've learned.

Thanks

• I'm not expert in formal logic at all, but P2 and P3 really bother me. Don't they imply $L\rightarrow\neg L$? May 21, 2015 at 22:13
• but they in the questions are : P2) L → D P3) D → ¬L May 21, 2015 at 22:20
• Hmmm, well, like I said, I'm not an expert. Good luck! May 21, 2015 at 22:22
• @bob.sacamento That we a mistake from me (texing error) apologies for to you and the OP. May 21, 2015 at 22:24
• @user155971 to make up for that error, from P2 and p3 you can get $\neg L$ (as if $L$ is true, then so is $\neg L$, a contradiction). From P1 you get $E\lor(L\lor M)$ which is the same as $L\lor (E\lor M)$, and from this and $\neg L$ you get $(E\lor M)$. How that all fits in your proof system depends on the details of your proof system. May 21, 2015 at 22:26

We have the following deduction:

1) $L\rightarrow(\lnot L)$ by hypothetical syllogism and P2,P3.

2) $(\lnot L)\vee(\lnot L)$ by material implication and 1.

3) $\lnot L$ by disjunctive tautology and 2.

4) $E\vee (L\vee M)$ by conjunctive simplification and P1.

5) $(E\vee L)\vee M$ by disjunctive associativity and 4.

6) $(L\vee E)\vee M$ by disjunctive commutativity and 5.

7) $L\vee (E\vee M)$ by disjunctive associativity and 6.

8) $E\vee M$ by disjunctive syllogism and 7,3.

Conclude that the argument is valid.

• thank you so much , but can you recommended me with a good book explain logic May 21, 2015 at 23:00
• It depends on the level at which you'd like to study logic, the topics you want to study, and the ways you'd eventually like to use it. If you're just looking to do this stuff (it's called propositional calculus), then just about any text will do as long as it has plenty of exercises. Practice is the only way to get better at natural deduction... I'd bet you can find plenty of exercises through Google too. May 22, 2015 at 0:42
• $$(E \lor (L \lor M)) \land (E \iff F)\equiv(E \lor (L \lor M))\land (E\implies F\land F\implies E)$$ $$\equiv(E \lor L \lor M)\land\neg(\neg(E\implies F)\lor\neg(F\implies E))$$ $$\equiv(E \lor L \lor M)\land\neg((E\land\neg F)\lor(F\land\neg E))$$ $$\equiv(E \lor L \lor M)\land(\neg E\lor F)\land(E\lor \neg F)\equiv(E \lor L \lor M)\land(E\lor \neg F)\land(\neg E\lor F)$$ $$\equiv (E\lor((L\lor M)\land \neg F))\land (\neg E\lor F)$$$$\equiv(((E\lor((L\lor M)\land \neg F))\land \neg E)\lor(((E\lor((L\lor M)\land \neg F))\land F)$$ $$\equiv((L\lor M)\land\neg(E\lor F))\lor(E\land F)$$ Dec 10, 2019 at 16:25
• If it means anything Dec 10, 2019 at 16:29

Here is a proof using the proof checker from the forallx text. The OP in a comment is looking for recommended books explaining logic. This text with the associated truth functional and first order logic proof checker may supplement what the OP is currently using.

Here is the proof:

Here is a summary of the proof:

• Use conjunction elimination (∧E) to get the first conjunction from premise 1. We will not need the second conjunct from that premise.
• Since this conjunct is a disjunction consider both cases of the disjunction separately. If we can reach the desired conclusion in both cases then we can eliminate the disjunction.
• The first case $$E$$ is handled from lines 5 to 6. Notice the disjunction introduction (vI) on line 6.
• The second case $$L \vee M$$ is handled from lines 7 to 14. It requires handling another disjunction by considering both cases to reach the desired goal.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/