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I'm trying to solve this(Classical Propositional Logic) :

$$ \gamma \ \lor ( \lnot \ \gamma \ \lor (\ \psi \land \ \phi \ )) \, $$

I did this :

  1. Assuming $$\psi,\phi$$
  2. Introduction the $$\land$$ so $$(\ \psi \land \ \phi \ )$$
  3. Now I can use the introduction of $$\lor$$ so $$\gamma \ \lor (\ \psi \land \ \phi \ )$$ 4).And now?

Sorry I my first post, and also I'm new in logical deduction, so please halp me and don't rate me wrong.
P.S I use Tree Proofs

here

1 Answers 1

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Let $A = (\gamma \ \lor ( \lnot \ \gamma \ \lor (\ \psi \land \ \phi \ ))$. We want to prove $\vdash A$.

Assume $\neg A$ and $\gamma$.

From $\gamma$, by $\lor I$ we have $(\gamma \ \lor ( \lnot \ \gamma \ \lor (\ \psi \land \ \phi \ ))$, then by $\neg E$ we get $\bot$.

Now from $\bot$ using $\neg I$ we can discharge $\gamma$ and deduce $\neg \gamma$.

From $\neg \gamma$ using $\lor I$ we get $\lnot \ \gamma \ \lor (\ \psi \land \ \phi \ )$.

Using $\lor I$ again we can deduce $A$, and from $A$ and $\neg A$, using $\bot$ law, we get $\bot$.

From $\bot$, using $RAA$ we get $A$ and discharge $\neg A$, and the proof is complete.

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    Can you show me another way, without using the $\bot $ and also all the family A, just piece per piece? – 2017-02-02
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    This can not be proved without using the $\bot$ If you want to prove by natural deduction since $\vdash \gamma \lor \neg \gamma$ is not one of the rules of natural deduction and you have to prove it. – 2017-02-02
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    Also, you can replace $A$ with $(\gamma \ \lor ( \lnot \ \gamma \ \lor (\ \psi \land \ \phi \ ))$ throughout the proof. I've written every step of the proof with detail. – 2017-02-02
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    Ok, sorry I'm really noob, can you give me some ref, site or something else to study it ? – 2017-02-02
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    You can find a very detailed proof of LEM and some references here: http://math.stackexchange.com/questions/2054315/proving-law-of-excluded-middle-in-fitch-system/2054347 – 2017-02-02